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### Application of Exponential and Logarithmic Functions

• Exponential and logarithmic functions have a huge array of applications. They are used in science, business, medicine, and even more fields. There are far too many applications to discuss them all in this lesson, so instead this lesson focuses on working a variety of examples.
• Exponential functions allow us to describe the growth (or decay) of a quantity whose rate of change is related to its current value. Here are some examples of applications:
• Compound interest,
• Depreciation (loss in value),
• Population growth,
• Half-life (decay of radioactive isotopes),
• Any many others!
• Many exponential functions have their own specific formulas, however, if you forget any of those formulas, it is sometimes possible to use the natural exponential growth model:
 Pert.
It will not use the same rate r as if you had used the specialized formula, but if you can figure out what the r needs to be for Pert from the problem, you can normally get away with using it instead.
• Logarithms have the ability to capture the information of a wide variety of inputs in a relatively small range of outputs. This behavior makes logarithms a great way to measure quantities that can be vastly different, but need an easy way to be compared and talked about:
• Earthquake magnitude (Richter scale),
• Sound intensity (decibels),
• Acid/base concentration (pH scale),
• And many others!
Logarithms also show up in formulas that analyze how exponential growth behaves.
• Carbon-14 is a radioactive isotope with a half-life of 5 730 years. It is created in the upper atmosphere by cosmic rays, and is then absorbed by living organisms (plants and animals). Carbon-14 makes up a small, but consistent, amount of the carbon in any live organism. However, once the organism dies, C-14 stops being absorbed. Without the isotope being replenished, the quantity of C-14 in a dead organism begins to decline. This allows for carbon-14 dating (AKA radiocarbon dating, carbon dating). By knowing how much C-14 would be in a live organism, then measuring the amount in a dead organism, we can do archeological dating based on how much C-14 remains.
• We can figure out how quickly something will heat up or cool down with Newton's Law of Cooling. The rate at which heat is transferred is proportional to the difference in temperature between the object and the surrounding environment. This results in the equation
 T = Ts + (Ti − Ts) ekt,
where T is the object's temperature, Ts is the surrounding environment, Ti is the object's initial temperature, k is a proportionality constant, and t is the time.

### Application of Exponential and Logarithmic Functions

A bank account that compounds continuously is opened with an original principal of \$15 000 placed in it. After 10 years from the account's opening, it has a balance of \$25 484. What will its balance be after 20 years from the account's opening?
• Note: All of the solution steps for the questions in this section assume a high degree of familiarity with solving exponential and logarithmic equations. While the set up of problems will be explained, there will not usually be a detailed explanation of the steps involved in solving the equations once set up. If you are not currently comfortable with solving such equations, make sure to watch the previous lesson, Solving Exponential & Logarithmic Equations.
• The amount A in a bank account with continuous compounding is modeled with the same equation as anything with continuous growth (or decay):
 A = Pert,
where P is the starting quantity, r is the rate coefficient, and t is the time elapsed.
• Notice that we do not know the rate r, so we cannot immediately plug in t=20 to find the balance after 20 years. However, we can plug in the information we know at t=10, then solve for r:
 25484 = 15000er·10
• Solve for r:
e10r = 25484

15000
⇒     10r = ln
25484

15000

⇒     r = ln ⎛⎝ 25484 15000 ⎞⎠

10
Plugging in to a calculator, we find that r = 0.053.
• Now that we know r, we have enough information to see what the balance is at t=20:
 A = 15000 e0.053 ·20     =     43 295.56
\$43 295.56
A bank account compounds every quarter, and is initially opened with a principal of \$22 000. After seven years, it has a balance of \$28 273. What is the balance of the account 15 years after the account is opened? (Round to the nearest whole dollar.)
• Note: All of the solution steps for the questions in this section assume a high degree of familiarity with solving exponential and logarithmic equations. While the set up of problems will be explained, there will not usually be a detailed explanation of the steps involved in solving the equations once set up. If you are not currently comfortable with solving such equations, make sure to watch the previous lesson, Solving Exponential & Logarithmic Equations.
• From previous lessons, we know that for bank accounts that do not compound continuously, we can use the equation
 A = P ⎛⎝ 1 + r n ⎞⎠ nt .
Because the account compounds quarterly, we have n=4. To find the value at t=15, we first need to find the rate r by using what we know at t=7.
• Setting it up at t=7, we have:
 28273 = 22000 ⎛⎝ 1 + r 4 ⎞⎠ 4·7
Now we work towards solving for r (Notice how we take a root, not a logarithm. This is because we're interested in the base, not solving for the exponent.):

1 + r

4

28

= 28273

22000
⇒
1 + r

4

=
28

 28273 22000

⇒     r =

28

 28273 22000

−1

·4
Plugging in to a calculator, we find r = 0.036
• Now that we know r, we can find the balance at t=15:
 A = 22000 ⎛⎝ 1 + 0.036 4 ⎞⎠ 4·15 ≈     37 661
• Alternatively, we could solve the problem in another way. In the lesson, it is mentioned that we can often use the exponential growth model to replace many other formulas:
 A = Pert
We can do this here as well (see the lesson for an explanation of why we are able to do this).
• Setting up with this formula, we can solve for r using information about the balance at t=7:
28273 = 22000 er·7     ⇒     r·7 = ln
28273

22000

⇒     r = ln ⎛⎝ 28273 22000 ⎞⎠

7
This gives us r=0.035 838. [Notice how this is a different value for r then the one we found when using the non-continuous compounding formula. Anytime you choose to use Pert instead of the standard formula, you will find a different value for your r rate.]
• Now we can find the balance at t=15:
 A = 22000 e0.035838 ·15    ≈     37 661
This goes to show that in many situations where exponential growth is involved, we can often just default to using Pert if we so desire.
\$37 661
A bank account that compounds continuously is opened with an original principal of \$7 000 placed in it. After 20 years from the account's opening, it has a balance of \$19 400. What is the percentage rate of the account given to three decimal places?
• Note: All of the solution steps for the questions in this section assume a high degree of familiarity with solving exponential and logarithmic equations. While the set up of problems will be explained, there will not usually be a detailed explanation of the steps involved in solving the equations once set up. If you are not currently comfortable with solving such equations, make sure to watch the previous lesson, Solving Exponential & Logarithmic Equations.
• The amount A in a bank account with continuous compounding is modeled with the same equation as anything with continuous growth (or decay):
 A = Pert,
where P is the starting quantity, r is the rate coefficient, and t is the time elapsed.
• For this problem, we are looking to solve for r. Set up the equation:
 19400 = 7000er·20
• Solve for r:
e20r = 19400

7000
⇒     20 r = ln
19400

7000

⇒     r = ln ⎛⎝ 19400 7000 ⎞⎠

20
Plugging into a calculator, we find r = 0.050 968.
• We aren't quite done, though. The problem asked for r as a percentage rate, not a decimal coefficient. We convert it to a percentage by shifting the decimal place over two positions:
 r = 0.050 968     ⇒     5.0968%
Finally, the problem asked for the percentage to three decimal places, so we must round to three decimal places: 5.097%.
5.097%
A bank account that compounds continuously is opened with some original principal placed in it. After 10 years, the account has \$1858 in it. After a further 10 years, it has \$3456 in it. What was the original principal in the account, rounded to the nearest whole dollar?
• Note: All of the solution steps for the questions in this section assume a high degree of familiarity with solving exponential and logarithmic equations. While the set up of problems will be explained, there will not usually be a detailed explanation of the steps involved in solving the equations once set up. If you are not currently comfortable with solving such equations, make sure to watch the previous lesson, Solving Exponential & Logarithmic Equations.
• The amount A in a bank account with continuous compounding is modeled with the same equation as anything with continuous growth (or decay):
 A = Pert,
where P is the starting quantity, r is the rate coefficient, and t is the time elapsed.
• Notice that we do not know P or r currently. However, we know the account balance at t=10 and t=20, allowing us to set up the equations below:
 1858 = P er·10               3456 = Per·20
From here, we can work to solve through substitution.
• Working with the first of the above equations, we have
 P = 1858 e10r .
We can now plug this in to the other equation:
 ⎛⎝ 1858 e10r ⎞⎠ e20r = 3456
• From here, we work to solve for r:

1858

e10r

e20r = 3456     ⇒     1858 e10r = 3456     ⇒     10r = ln
3456

1858

⇒     r = ln ⎛⎝ 3456 1858 ⎞⎠

10
Plugging into a calculator, we find r=0.062 061.
• Now that we know r, we can plug it into our previous equation to find the value of P:
 P = 1858 e10·0.062061 ≈     999
\$999
At a lab, a population of bacteria is growing exponentially. A scientist checks the bacteria culture at 8 and finds that there are 5 000 bactera in the culture. At 1, he checks again and finds that there are 417 000 bacteria. How many bacteria will be in the culture when the scientist leaves the lab at 5? (Round to the nearest thousand.)
• Note: All of the solution steps for the questions in this section assume a high degree of familiarity with solving exponential and logarithmic equations. While the set up of problems will be explained, there will not usually be a detailed explanation of the steps involved in solving the equations once set up. If you are not currently comfortable with solving such equations, make sure to watch the previous lesson, Solving Exponential & Logarithmic Equations.
• Because the population of bacteria grows exponentially, we can model it with the equation
 A = Pert.
• We can treat the first time the scientist checks the culture as t=0. Thus, the population at t=0 tells us our initial population: P=5 000. Next, we need to find the rate r of growth from the next piece of information.
• Since the scientist checks the culture again five hours later, we can consider that t=5. This allows us to set up the below equation:
 417000 = 5000 er ·5
• Solve for r:
e5r = 417000

5000
⇒     5r = ln
417

5

⇒     r = ln ⎛⎝ 417 5 ⎞⎠

5
Using a calculator, we get r=0.884 730.
• Finally, we want to know the population at 5, which is nine hours after beginning, so t=9:
 A = 5000 e0.884730 ·9     =     14 357  215
We were told to round to the nearest thousand, so we get a population of 14 357  000.
14 357  000 bacteria
Plutonium-238 is a radioactive isotope with a half-life of 87.7 years. If you start with 1kg of Pu-238, how long will it take to decay down to 1g? (Round to the nearest year.)
• Note: All of the solution steps for the questions in this section assume a high degree of familiarity with solving exponential and logarithmic equations. While the set up of problems will be explained, there will not usually be a detailed explanation of the steps involved in solving the equations once set up. If you are not currently comfortable with solving such equations, make sure to watch the previous lesson, Solving Exponential & Logarithmic Equations.
• First, notice that in the problem uses two different units for mass: kilograms (kg) and grams (g). Thus, before we get started, we want to convert to using just a single unit. Let's convert the kilograms: we start with 1000g of Pu-238, and we end up with 1g.
• We can model the amount of undecayed Pu-238 as
 A = 1000 ⎛⎝ 1 2 ⎞⎠ [t/87.7] ,
where t is the number of years since we started with the mass of Pu-238. [We could also model this with Pert  ⇒  1000 ert, but the downside is that we would have to begin by figuring out what r for the natural base (e) is by using the half-life value of 87.7 years: 500 = er·87.7 If instead we do it with the first equation given above, we don't have to figure out a rate coefficient, so it makes the problem slightly easier. But the method using Pert will work perfectly fine, it just requires an extra step. (Notice that the r coefficient is very different when we work with e as the base).]
• We want to know what t is when the amount is A=1, so we plug in and solve:
 1000 ⎛⎝ 1 2 ⎞⎠ [t/87.7] = 1     ⇒ ⎛⎝ 1 2 ⎞⎠ [t/87.7] = 1 1000 ⇒ t 87.7 log 1 2 = log 1 1000
[Remember, we can do the above because a logarithm of any base can be used to bring down an exponent. Thus base 10 or base e are good choices because they are easily accessible on most calculators.]
t

87.7
= log 1 1000

 log 1 2
⇒     t =

 log 1 1000

 log 1 2

·87.7
Using a calculator, we find t = 874.
874 years
Silicon-32 is a radioactive isotope that decays over time. A scientist begins with 1.000  000  g of Si-32. A year later, she measures the amount of Si-32 to find there is only 0.995 931 g of the isotope remaining. What is the half-life of Si-32? (Round to the nearest year.)
• Note: All of the solution steps for the questions in this section assume a high degree of familiarity with solving exponential and logarithmic equations. While the set up of problems will be explained, there will not usually be a detailed explanation of the steps involved in solving the equations once set up. If you are not currently comfortable with solving such equations, make sure to watch the previous lesson, Solving Exponential & Logarithmic Equations.
• We can model the amount of undecayed Si-32 as
 A = 1.000 000 · ⎛⎝ 1 2 ⎞⎠ [t/n] = ⎛⎝ 1 2 ⎞⎠ [t/n] ,
where t is the number of years since we started with the original mass of Si-32 and n is the half-life of the isotope.
• Thus, we can solve for n by plugging in the information for a year later:
0.995 931 =
1

2

[1/n]

⇒     log(0.995931) = 1

n
log 1

2
⇒     n = log 1 2

log0.995931
Plugging into a calculator, we find n=170.
• Alternatively, we could approach this problem using the Pert formula. First, we need to find the value of r:
 0.995931 = 1 ·er·1     ⇒     0.995931 = er    ⇒     ln0.995931 = r
Using a calculator, we find r = −0.004 077. Next, we need to figure the length of time for one half-life.
• This requires a little bit of clever thought. We must think about how much isotope would be left over at the time of one half-life. By definition, a half-life is the amount of time it takes for half of the material to decay, so we must have half as much material at the end of one half-life cycle. For this problem, since we started with 1g of material, at the end of one half-life we would have 0.5g of material. Let t represent the time of one half-life, and we will have the equation
 0.5 = 1·e−0.004077 ·t.
We can now solve this equation for t to find the half-life:
 e−0.004077 t = 0.5     ⇒     −0.004077 t = ln0.5     ⇒     t = ln0.5 −0.004077
Plugging into a calculator, we get our half-life time of t=170.
The half-life is 170 years.
Carbon-14 has a half-life of 5730 years and is present in living organisms at trace amounts and keeps a constant ratio with non-radioactive carbon while an organism is alive. A wooden spear is discovered in an archeological dig, and the wood has a C-14 ratio that is 0.34% of the ratio for living wood. Estimate the age of the spear to the nearest thousand years.
• Note: This problem is based on the idea of carbon-14 dating. If you're curious to understand how the mechanism of carbon-14 dating (AKA radiocarbon dating, carbon dating) works, check out the video lesson for more information. The idea is introduced as a prelude to Example 2.
• We can model the amount of undecayed C-14 as we have in the previous couple problems about isotopes, but we need to state how much C-14 we begin with when the tree dies and the spear is made. Since the spear is made just after the tree's death (or at least relatively soon, compared to the half-life of C-14), we can say that the spear had 100% of the ratio for living wood (because the C-14 has not had time to start decaying significantly). Thus, let's measure the amounts of C-14 with an arbitrary "points" system based on the percentages. When the spear is first made, it has 100 points. When the archeologists find it, it has 0.34 points.
• Thus, we have the following equation, where t is the age of the spear when the archeologists find it:
 0.34 = 100 · ⎛⎝ 1 2 ⎞⎠ [t/5730]
• Solve for t:

1

2

[t/5730]

= 0.34

100
⇒     t

5730
log 1

2
= log 0.34

100
⇒     t = log 0.34 100

 log 1 2
·5730
Plugging into our calculator, we find t ≈ 47 000.
• As a side note, it would be possible to do this problem using Pert, but it would take a little more effort. We would have to begin by figuring out what r is for the half-life of 5730, which would then allow us to solve for the age of the spear. Nonetheless, we will get the same answer either way, so feel free to use the method you are more comfortable with.
The spear is 47 000 years old.
A pot of soup is on the stove and has a temperature of 85°C when you turn off the stove. You immediately move the pot of soup to the refrigerator, which has a constant interior temperature of 3°C. After five minutes of it being in the refrigerator, you check the soup and see that its temperature is now 72°C. If you can only eat the soup once it reaches 55°C, how much longer must you let the soup cool in the refrigerator after checking it? (Round to the nearest minute.)
• Note: This problem is based on Newton's Law of Cooling, which allows us to model the temperature of an object based on its initial temperature and the temperature of its surroundings. The equation and the idea are introduced as a prelude to Examples 4 and 5 in the video lesson. The equation is
 T = Ts + (Ti − Ts) ekt,
where T is the object's temperature, Ts is the surrounding environment, Ti is the object's initial temperature, k is a proportionality constant we figure out for each problem, and t is the time.
• The initial temperature of the soup is Ti = 85 and the surround temperature in the refrigerator is Ts=3. At t=5, we have the equation
 72 = 3 + (85−3) ek·5,
which we can now solve for the coefficient k.
• Solving for k, we get
3 + 82 e5k = 72     ⇒     e5k = 69

82
⇒     5k = ln 69

82
⇒     k = ln 69 82

5
Using a calculator, we find k = −0.034 523.
• Now that we know the value for k, we can set up the equation for an unknown t when the temperature of the soup is at 55°C:
 55 = 3 + (85−3) e−0.034523·t
• Now we solve for t:
3 + 82e−0.034523t = 55     ⇒     e−0.034523t = 52

82
⇒     t = ln 52 82

−0.034523
Plugging into a calculator, we find that the soup is cool enough at t ≈ 13.
• However, the question asked for how much longer the soup had to cool. This was compared to when the temperature was checked at t=5. Thus, the amount of time longer is 13−5   =  8 minutes.
You must let the soup cool for another 8 minutes. [Notice that it is 8 minutes after checking the temperature in the refrigerator, which comes to a total of 13 minutes after being initially taken off the stove.]
During a cold winter night, the police get a call about a disturbance in an alleyway. They arrive on the scene a little while later, and find a body. They take the temperature of the body at 10:51, and it is 31.5°C. They proceed to secure the area and call a detective to the scene. The detective arrives at 11:10 and she immediately takes the temperature of the body, which is now at 27.3°C. She also takes the temperature of the alleyway, which has been holding at a constant −15°C for the past couple hours. Assuming that the victim had a standard human body temperature of 37°C, give the victim's time of death (to the nearest minute).
• Note: This problem is based on Newton's Law of Cooling, which allows us to model the temperature of an object based on its initial temperature and the temperature of its surroundings. The equation and the idea are introduced as a prelude to Examples 4 and 5 in the video lesson. The equation is
 T = Ts + (Ti − Ts) ekt,
where T is the object's temperature, Ts is the surrounding environment, Ti is the object's initial temperature, k is a proportionality constant we figure out for each problem, and t is the time.
• (It should be noted that this is a particularly challenging problem. If you find it difficult to follow the steps below, consult the video lesson for Example 5, which works out a very similar problem.) Let's begin by establishing how we look at the problem. Let t=0 be the time of the victim's death and let the unit of time be minutes. Notice that we do not know the time that the police first measure the temperature of the body. Since this is an unknown, let us call it t0. Thus, the police measure the body's temperature at t = t0, and therefore the detective measures the body's temperature 19 minutes later at t = t0 + 19.
• Immediately at death, the body has an initial temperature of Ti = 37 and the surrounding environment is Ts = −15. Using Newton's Law of Cooling, the measuring done by the police and the measuring done by the detective each give us an equation to work with:
 31.5 = −15 + (37 − (−15)) ek t0            27.3 = −15 + (37 − (−15)) ek (t0+19)
Moving things around a bit in the equations, we get
 46.5 = 52ek t0            42.3 = 52 ekt0 + 19k
• At this point, we have two unknowns: t0 and k. Since we have two equations, we can use substitution to solve. Working towards this, notice that we can rewrite the second equation as
 42.3 = 52 ekt0 + 19k     ⇒     42.3 = 52 ( ekt0 ·e19k )     ⇒     42.3 = 52 ek t0 ·e19k
At this point, we can substitute in our first equation, 46.5 = 52ek t0:
 42.3 = 52 ek t0 ·e19k     ⇒     42.3 = (46.5) ·e19k
• From here, we can now solve for k:
46.5 e19k = 42.3     ⇒     e19k = 42.3

46.5
⇒     19k = ln 42.3

46.5
⇒     k = ln 42.3 46.5

19
Using a calculator, we find k = −0.004 982.
• We can now plug this in to the first of our two temperature equations to solve for t0:
46.5 = 52e−0.004982 t0     ⇒     −0.004982 t0 = ln 46.5

52
⇒     t0 = ln 46.5 52

−0.004982
Using a calculator, we find t0 = 22.439.
• The problem asked for the time of death to the nearest minute, so we round to t0≈ 22. Remember, t0 represents the number of minutes after the victim's death. We know that t0 occurs at 10:51. Since that is 22 minutes after the victim's death, we see that the victim's death (t=0) occured at 10:29. This completes the problem.
• The above is a very good, clear way to do the problem, and it is very similar to the method used in the video lesson. There is an interesting alternative way to do the problem, though. Instead of setting the victim's time of death as t=0, set the initial time of t=0 at the instant that the police first measure the body's temperature. This means we link 10:51 with t=0. Furthermore, because we've linked that moment with the initial time, it also means we link the body's temperature at that moment to the initial temperature: in other words, we set Ti = 31.5 (what the police measure at t=0↔ 10:51). From here, we can solve for the value of k using the detective's temperature measurement 19 minutes later at t=19:
 27.3 = −15 + (31.5 − (−15) ) ek·19
• Solve for k:
46.5 e19k = 42.3     ⇒     e19k = 42.3

46.5
⇒     19k = ln 42.3

46.5
⇒     k = ln 42.3 46.5

19
Plugging into a calculator, we get k = −0.004 982. (Notice that is the exact same value as we found the first time. This makes sense, because the coefficient k is just a mathematical expression of how quickly heat energy transfers between the object and its surrounding environment. It is based on how the object and the environment interact, not the specific temperatures involved.)
• Now that we know k, we can work out what time t the body would have had a normal (living) body temperature of 37°C:
 37 = −15 + (31.5 −(−15)) e−0.004982 t
Solving for t, we find
46.5 e−0.004982 t = 52     ⇒     −0.004982 t = ln 52

46.5
⇒     t = ln 52 46.5

−0.004982
Plugging into a calculator, we find that the time of death t=−22.439. Round to the nearest minute of t ≈ −22, and realize that this means the death occurred in "negative" time, that is prior to t=0. This makes perfect sense, since it's obvious that death must have occurred prior to the time the police measured the body's temperature. Working back 22 minutes from 10:51, we get a time of death at 10:29.

Both methods of solving the problem are perfectly valid. The important part is just keeping track of what moment in time you decide to link with the "initial" moment of t=0. We can say the initial moment is the moment of death, or the initial moment is when the police first check the body. The important thing isn't which one we choose, just that we keep it straight throughout working on the problem.
The victim died at 10:29. [If you're interested, there are two different methods to set up and solve the problem given in the steps for this question. The first part of the steps uses a method similar to the one in the video lesson, but after that is an alternative method where we look at the problem in a slightly different way.]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Application of Exponential and Logarithmic Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Introduction 0:06
• Applications of Exponential Functions 1:07
• A Secret! 2:17
• Natural Exponential Growth Model
• Figure out r
• A Secret!--Why Does It Work? 4:44
• e to the r Morphs
• Example
• Applications of Logarithmic Functions 8:32
• Examples
• What Logarithms are Useful For
• Example 1 11:29
• Example 2 15:30
• Example 3 26:22
• Example 4 32:05
• Example 5 39:19

### Transcription: Application of Exponential and Logarithmic Functions

Hi--welcome back to Educator.com.0000

Today we are going to talk about applications of exponential and logarithmic functions.0002

At this point, we have a good understanding of exponentiation and logarithms.0007

In this lesson, we will see some of the many applications that they have.0010

Exponential and logarithmic functions have a huge array of applications.0013

They are used in science, in business, in medicine, and even more fields.0017

They are used in all sorts of places.0021

There are far too many applications to discuss them all in this lesson, so instead we will focus on working a variety of examples.0024

We will begin with a brief overview of some other uses--some of the uses that we can see for exponential functions and logarithmic functions.0029

Then, we will look at many specific examples, so we can really get our hands dirty and see how word problems in this form work.0035

Now, before you watch this lesson, make sure you have an understanding0040

of exponents, logarithms, and how to solve equations involving both, before watching this.0043

We won't really be exploring why the actual nuts and bolts of this solving works--how these things work.0048

We are just going to be launching headfirst into some pretty complicated problems.0053

So, you really want to have an understanding of what is going on, because we are going to hit the ground running when we actually get to these examples.0057

Previous lessons will be really, really helpful here if you are not already used to this stuff.0062

OK, let's go: applications of exponential functions: exponential functions allow us to describe the growth or decay0066

of a quantity whose rate of change is related to its current value.0073

So, how fast it is changing is connected to what it is currently at.0078

So, some examples of applications: we can also see how their rate of change is related to its current value:0083

Compound interest--the amount of interest that an account earns is connected to how much money is already in the account.0088

If you have ten thousand dollars in an account, it will earn more than if it has one thousand dollars in the account,0096

or than if it has one hundred dollars in the account.0101

So, this is an example of seeing how the rate of change is related to the current value of the object.0103

Other things that we might see: depreciation--loss in value; compound interest and loss in value0109

are both used a lot in banking and business--anything that is fiscally oriented.0114

Population growth is used a lot in biology; half-life--the decay of radioactive isotopes--shows up a lot when we are talking about physics.0118

If you are studying anything in radiation, understanding half-life is very useful.0127

And many others--there is a whole bunch of stuff where exponential functions are going to show up.0131

It is really, really useful stuff.0135

Many exponential functions have their own formula--things like compound interest, 1 plus the number of times that it compounds in a year, divided...0142

Oh, this should actually be the other way around; it should not be n/r; it should r/n.0152

The rate of it, divided by the number of times it compounds, to the number of times it compounds, times the time--0160

if you don't remember that one, remember our very first lesson on exponential functions that described why that is the case.0167

Population doubling is P, some original starting principal amount, times 2 to the rate that they double at, times t.0173

Half-life is some principal starting amount times 1/2 to the rate times time.0181

However, if you forget all of these formulas--there are a bunch of different formulas;0187

there are even more than just these; but it is sometimes possible to use the natural exponential growth model.0191

You can sometimes swap out any one of these more difficult-to-remember ones for simple "Pert"--0197

P, the original starting amount, times e, the natural base, to the r times t,0205

where r is the rate of the specific thing that we are modeling, times time.0210

r will change, depending on what different thing you are doing.0214

So, even if you are modeling isotopes--half-life in plutonium and half-life in uranium--you will get very different r's,0217

because the plutonium and uranium will have different rates of decay.0222

So, you are not going to use the same rate r.0226

Once again, if you are talking about half-life, the r here would be totally different than the r in our Pert if you had used the specialized formula.0229

So, the r for half-life of uranium using this formula would be totally different than the r for half-life using the Pert formula.0236

But if you can figure out what that r has to be for Pert form from the problem, you can get away with using it instead.0246

There are many situations where you might not remember any one of these specialized formulas.0255

But it can be OK if you have enough information from the problem to be able to figure out what r has to be.0260

There are lots of cases where that will end up being the case.0266

We will talk about a specific one on Example 2; we will see something where we could get away with not knowing0269

the specific formula, and still be able to figure things out by using this Pert formula.0275

We will talk about it in Example 2, if you want to see a specific example of being able to use this secret trick.0279

Why is this the case?--you probably wonder.0285

Why can you do this--how can you get away with using Pert when we have all of these special formulas?0287

How can you swap out some exponential formula for just natural exponential growth, the Pert form?0291

The reason why is because of this er part; er can morph into other forms.0297

For example, let's look specifically at a possible half-life formula.0303

We might have P times 1/2 to the t/5; we can see this as P times 1/2 to the rate of 1/5 times t.0307

That is what we have there: some principal starting amount, times 1/2 to the 1/5 times t.0318

So, for every, say, 5 years, we have half of the amount there that we originally had.0323

So, how can we get Pert to connect to this?0328

Well, if we use this very specific r, r = -0.1386, it also turns out that that is the same thing as -0.6931 times 1/5.0331

So, we can have our Pert form right here; we know what r is, so we swap that in for our r.0343

And we get P times e to the -0.1386 times time.0350

But we also know that -0.1386 is the same thing as -0.6931 times 1/5.0358

So, if we want, we can break this apart into a -0.6931 part and a 1/5 times t part that we might as well put outside.0366

We have e to the -0.6931, to the 1/5 times t, because by our rules from exponential properties, that is the same thing0374

as just having the 1/5 and the -0.6931 together, which is the r that we originally started with.0381

Now, it turns out that e-0.6931 comes out to be 1/2.0387

By this careful choice of r, we are able to get er to morph into something else.0396

We can get it to morph into this original 1/2; and now, we have this 1/5 here,0402

so it becomes just P times 1/2 to the t/5, which is what we originally started with as the half-life formula.0406

So, by this careful choice of r--and notice, the r here is equal to 1/5; the r here is equal to the very different 0.1386;0414

we get totally different r's here; but by choosing r carefully, if we have enough information from the problem0426

(sometimes you will; sometimes you won't; you will have to know that special formula)--sometimes,0433

you will be able to get enough information from the formula, and you will be able to figure out what r is.0438

So, you can have forgotten the special formula--you can forget the special formula occasionally, when you are lucky.0441

And you would be able to just use Pert instead.0446

By choosing the right r for Pert, we morph it into something that works the same as the other formula.0448

Now, of course, you do have to figure out the appropriate r from the problem.0454

You are just saying it--you have to be able to get what that r is.0457

And remember: it is the r for Pert, which may be (and probably is) going to be totally different0460

than the r for the special formula that we would use for that kind of problem.0465

But if you can figure out what the r is from the problem, you can end up using Pert instead.0469

Once again, we will talk about a specific use of this in Example 2, where we will show how you can actually use this if you end up forgetting the formula.0473

Now, I want you to know that the above isn't precisely true.0480

e-0.6931 isn't precisely 1/2; it is actually .500023, which is really, really, really close to 1/2; but it is not exactly 1/2.0484

But it is a really close approximation, and it is normally going to do fine for most problems.0496

It is such a close approximation that it will normally end up working.0500

And if you need even more accuracy, you could have ended up figuring out what r is, just to more decimal places.0504

And you could have used this more accurate value for r.0509

Applications of logarithmic functions: logarithms have the ability to capture the information of a wide variety of inputs in a relatively small range of outputs.0513

Consider the common logarithm, base 10: if we have log(x) equaling y, log(x) going to y--0522

over here we have our input, which is the x, and our output, which is the value y--that is what is coming out of log(x).0528

x can vary anywhere from 1 to 10 billion; and our output will only vary between 0 and 10.0537

That is really, really tiny variance in our output, but massive variance in our input.0545

Why is this happening? Because 1 is the same thing as 100,0550

and 10 billion is the same thing as 1010,because we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 zeroes,0555

so that is the same thing as 100 to 1010.0568

And because it is log base 10--it is the common logarithm--when log base 10 operates on 100, we will get 0.0571

When it operates on 1010, we will get 10; and as it operates on everything in between, we will get everything in between, as well.0577

So, there is massive variance in our inputs and massive different possible inputs that we can put in.0585

There is a very, very small range of outputs that we will end up getting out of it.0589

This behavior makes logarithms a great way to measure quantities that can be vastly different--0594

things that can have really huge variance in what you are measuring.0599

But we want an easy way to compare or talk about them; we have to be able to talk about these things.0602

They come up regularly, and we don't want to have to say numbers like 10 billion or 9 billion 572 million.0607

We want some number that is fairly compact--that doesn't require all of this talking.0614

So, we use logarithms to turn it into this much smaller, more manageable number that makes sense, and we can understand, relative to these other things.0618

Earthquake magnitude is one of the things that is measured on it.0625

It is measured on the Richter scale, which is a logarithmic scale.0629

Sound intensity is another one; it is measured in decibels, which is another logarithmic scale.0632

Acid or base concentration is measured on the pH scale; we will actually have examples about that in Example 3.0637

And that one is measured, once again, on a logarithmic scale.0643

And many others--there are many other logarithmic scales,0646

when you have a really, really large pool of information that can be going in as an input,0649

but you want to be able to narrow that to a fairly small, manageable, sensible range of values.0653

0 to 10 is going to have lots of decimals, when you evaluate log of 8 billion and 72 million.0658

It is going to have lots of possible decimals to it, but it is going to be a fairly small, manageable number for thinking about.0664

Logarithms will also show up in formulas that are analyzing exponential growth,0670

because if we are building a formula that is going to be connected to exponential growth,0675

if we are trying to break down and figure out what its power is raised,0678

we are going to end up having logarithms show up when we are solving it;0682

so they will end up showing up in the formula, as well.0684

Logarithms show up in formulas for analyzing exponential growth.0686

All right, let's get to some examples: A principal investment of \$4700 is made in an account that compounds quarterly.0690

If no further money is deposited, and the account is worth \$5457.57 in 5 years, what will it be worth after a total of 10 years?0698

The first thing: what kind of formula are we working with here?0707

Well, we have compounding, but not continuous compounding; so we go and look that up.0710

It is principal, times 1 + the rate, divided by the number of times that compounding occurs,0714

raised to the number of times that the compounding occurs, times the amount of time elapsed in years.0721

So, our principal investment here is P = 4700; and we know that, at time 5, at t = 5, we have 5457 dollars and 57 cents.0727

So, \$5457.57 is equal to...what was our principal amount? 4700 dollars, times 1 plus...what is our rate?0743

That is one of the things we don't know yet--we don't know what our rate is.0754

And that is why we are setting up at t = 5, instead of just hopping immediately to the 10 years question:0758

we need to figure out what our rate is first, so that is what we are figuring out now.0763

1 + r/n; our n is quarterly, so that is an n of 4, because it happens four times in a year, in each of the four quarters of the year.0767

So, 1 + r/4, raised to the 4 times t--do we know what t is in this case?0778

We do know what t is; otherwise we would have two unknowns to solve in one equation.0785

It is 5, because we are looking at the 5-year mark.0789

So, we work this out; we get 5457 dollars and 57 cents, over 4700 dollars, equals (1 + r/4)20.0793

Now, at this point, we would probably be tempted--how can we use logs?--how are logs connected to this?0806

We could take the natural log of both sides and bring down the 20; but then we would end up having ln(1 + r/4).0810

We are actually losing sight of a much simpler way.0816

When we raise to a power, how do we get rid of powers?0819

If we are trying to figure out what is in the base, and the power is a known value, we just take a root.0822

We take a root, depending on what the power is.0827

If it is squared, we take the square root; in this case, it is to the 20th, so we are going to raise both sides to the 1/20.0829

We are taking the twentieth root of this, so it is raised to the 1/20.0836

That cancels out here, and we are left with (1 + r/4); we take the 1/20 power, which is also the twentieth root,0842

of 5457 dollars and 57 cents, divided by 4700; and that comes out to be 1.0075.0855

We subtract 1 on both sides; we get 0.0075 = r/4.0864

We multiply by 4 on both sides, and we finally get 0.03 equals our rate.0869

So, our rate, in this case, is a modest 3% return on investment.0875

Now, we can use this information to look at the ten years, at t = 10.0881

At this point, we have everything that we need to know, except for the amount.0886

The amount is the unknown--we don't know what it is going to be worth.0889

But we do know what the initial principal was, \$4700; we do know that it is 1 + a rate of 0.03/4, quarterly,0892

to the 4 times...our number of years is 10; so our amount is 4700(1 + 0.03/4)40.0902

We work that all out with a calculator, and it ends up coming out to 6337 dollars and 24 cents.0914

And that is the final amount in the account at the 10-year mark.0925

The second example: carbon-14 dating: first, we are going to talk about the idea, and then we will actually work on a specific problem.0931

Carbon-14 is a radioactive isotope; a radioactive isotope is an element that is an isotope,0937

one version of an element, that breaks down over time; it decays into something else.0944

It is something for a while, and then something happens; it undergoes fission; it breaks apart; and it turns into a different element.0949

It is a radioactive isotope with a half-life of 5,730 years.0964

It is created in the upper atmosphere by cosmic rays.0968

Cosmic rays from the sun or other stars hit the upper atmosphere of the earth, and they create carbon-14 isotopes.0971

Then, these carbon-14 isotopes are absorbed by living organisms--that is, plants and animals.0982

They get absorbed by trees through the carbon cycle, and then the carbon ends up getting into the animals that eat those trees,0990

and the animals that eat those animals that eat those trees.0998

Any plant that is absorbing carbon will end up absorbing this, so it ends up getting into the cycle of biology.1001

Now, carbon-14--there is a very small amount of this isotope, so it is going to make up a small, but consistent, amount of the carbon of any live organism.1010

So, it is a very, very tiny amount, but it is going to be a regular amount.1018

As long as an organism is alive, it is going to continue consuming things.1022

It is going to continue bringing carbon-14 into its body; so that amount will stay basically consistent throughout its life.1027

However, once the organism dies, carbon-14 stops being absorbed.1033

There will be no more carbon-14 pulled in, because now, since the creature is dead,1039

since the organism is dead, it is not pulling any more carbon-14 into its body.1042

It is not pulling anything else into its body.1046

Without the isotope being replenished...remember, carbon-14 is radioactive; it begins to break down over time.1048

So, the C-14 (another name for carbon-14) in our dead organism will begin to decline,1054

because the creature is no longer pulling in carbon-14 to keep its levels consistent.1060

So, the carbon-14 begins to slowly, slowly drip away; it begins to slowly dissipate, break down, and decay into other things.1064

This fact allows for carbon-14 dating, also known as radiocarbon dating (because it is based on radioactive carbon), or simply carbon dating.1072

By knowing how much C-14 would be in a live organism (we know how much that is in a live organism,1082

because we can just measure it on real, live organisms--we can figure out that this creature is alive;1088

it has this much C-14 in it--a live creature, something that was very recently alive1093

or is currently alive--has this consistent amount of C-14), then once the creature dies, it begins to steadily decline.1099

So, we can measure the amount in a dead organism.1106

And since it is steadily declining--it is declining by rules that we know about--it is declining by half-life rules--1109

we can figure out, based on the amount of carbon-14 left in the dead organism at this point, when the creature died--1115

when it went from being alive and keeping a steady state of carbon-14 to dead,1124

where it is starting to let its carbon-14 just sort of disappear.1128

So, by knowing how much carbon-14 remains, we can get an estimate of when the creature was originally alive.1132

All right, now we will look at a specific example.1139

A skeleton of an ancient human is discovered during an archaeological dig.1141

If the level of carbon-14 in the bones is 7% of the carbon-14 ratio present in living organisms, how long ago did the human die?1145

And we want to round our answer to the nearest 1000 years.1152

And we are also reminded that the carbon-14 half-life is 5,730 years.1154

It is the amount that we originally start with, times 1/2 to the rate times time.1162

Now, in this case, the rate is going to be based on this 5730 years.1168

So, our amount is going to be equal to P--whatever our original amount is that we have of carbon-14--1172

to the 1/2, and we want a 1/2 to occur every 5730 years, so it is t/5730 years.1179

That way, when t = 5731, we will have an exponent of 1, and it will cause the 1/2 to go once.1191

If it is 5730 years twice, we will have an exponent of 2, and we will get 1/2 times 1/2: 1/4.1196

Great; so it makes sense what is going on there.1203

Now, what is our P going to be? We probably want to figure out what the P that we are going to use is.1205

Well, notice: it said the level of carbon-14 in the bones is 7% of the carbon-14 ratio.1210

So, we were never given absolute values; we weren't given quantities of carbon-14; we were told ratios.1215

If that is the case, well, what would we want the amount to be when it is 0--when we start at 0--when the creature was very first alive?1220

a = P(1/2)0/5730; well, that is going to be a number raised to the 0, which just becomes 1.1229

So, we have a = P; the amount that we start with is equal to the principal amount (that is exactly what it is there for).1237

If the ratio slowly declines, let's just say that it started at a ratio of 1.1246

We will say P = 1; this is an idea that we are figuring out.1251

We are saying, "Well, what would the ratio in a living organism be to a living organism? That would be 1, right?"1256

So, just before the creature dies at 0 time, it is going to have a ratio of living organism to living organism,1261

so it will be a ratio of 1; the ratio will not have changed.1268

So, initially, this P thing is just going to be 1, because the ratio at the instant of death is just 1,1272

because it is the normal thing with other living organisms.1278

So, our amount is 1/2 raised to the t/5730.1280

Now, at this point, we want to figure out what is the time that it took to get to where we are now.1286

So, at t = ? (we don't know what the time is), when we have it at 7%, we know that the amount is going to be 0.07.1293

The ratio if it is 7% is .07 with what it originally was: (1/2)t/5730.1305

Now, at this point, we could take log base 1/2, but I just feel like taking the natural log of both sides,1315

because we are going to have to do a change of base later anyway.1320

So, we have the natural log of 0.07 equals the natural log of 1/2 to the t/5730.1323

So, we can bring this down out front; we have t/5730.1333

ln(0.07) will divide by ln(1/2) on both sides; that equals t/5730, which finally brings us to t.1340

We multiply both sides by 5730; that equals 5730 times ln(0.07)/ln(1/2).1352

We punch that into a calculator, and we are going to end up getting about 21,983 as our time, which we round to 22,000 years.1363

This stuff isn't perfectly accurate, because...1374

So, the creature died 22,000 years ago; great.1378

This stuff can't be perfectly accurate, because there is always going to be a little bit of error in measurement.1382

And since we only had 7%, that is not as accurate as 21,983; so it makes sense that we can only get to a certain amount.1386

And that is why we are being told to round to the nearest 1000 years, because we only have so much specific data here.1393

Alternatively, if we had forgotten this a = P(1/2)rt, we could have started at the Pert form.1399

So, the amount equals Pert; once again, it is the same idea, this P = 1 business.1408

Let's start at a ratio of 1; we can plug that in, so we have a = ert.1414

Now, we need to figure out what r is; r is not going to be connected just directly to this 5730 years.1418

Instead, what we know is that, originally, at time = 0, we have 1 as the ratio.1425

At time = 5730, we are going to have 1/2 as the ratio.1436

And that is our trick here: 1/2 = er(5730).1441

We can take the natural log of both sides, so we have ln(1/2) = r(5730).1448

We divide both sides by 5730; we have ln(1/2)/5730 = r, which ends up getting us a rate of -0.00012097.1455

Now, that is long; it is a little complex; but it does mean that, even in the bad situation1475

where we have forgotten the formula for half-life--how half-life works--we can still get something that is going to work.1479

So now, we have this new thing we can work with, a = e-0.00012097t.1484

So, with that in mind, we know that we can look at 0.07 (put a little marker/divider down here);1497

0.07 = e to the same exponent, 0.00012097t.1505

We take the natural log of both sides; we have the natural log of 0.07 equals -0.00012097t,1513

because the exponent just comes down; we take the natural log.1524

So, now we have to divide this natural log of 0.07 divided by this huge...well, actually very tiny, but very long number, 0.00012097, equals t.1527

We punch that into a calculator, and we end up getting that t is approximately equal to 21,982,1540

which is very, very, very close to 21,983; so it ends up making sense that the reason why it is not quite the same1550

is because we ended up having to use a decimal approximation.1559

r wasn't exactly equal to -0.00012097; it was really close--it was approximately equal to that value.1562

So, we got something that was approximately the same thing as our answer.1571

And in this case, we are still going to end up rounding to 22,000 years as our final answer; cool.1574

The third example: first, let's talk about the pH scale.1583

The acid or base level of a solution (acid and base are two opposite ideas) is the concentration of positive hydrogen ions that are in solution.1585

This concentration is measured in moles per liter (moles is just a way of measuring how many of an atomic object you have), which is called molarity.1597

And it can vary hugely from one solution to another, so it makes sense that we want to use a logarithmic scale, since it can vary hugely.1606

To describe these massive possible differences, we measure acidity or alkalinity,1613

(acidity being the name for an acid measurement--how acid something is;1617

and base, basic, alkalinity, alkaline--these are all the same thing, for how basic something is--how "base" is a solution?)1622

with a logarithmic scale, the pH scale, which comes from powers of hydrogen,1630

which is connected to the fact that we are looking at a logarithm of our amount of hydrogen ions.1635

If we denote the molarity of hydrogen ions with the symbol H+, the pH is pH = negative log (common log,1640

so it is a base 10 log) of the molarity of hydrogen ions.1648

OK, with this idea in mind, we can now look at some interesting things.1653

The pH of cow's milk is 6.6, while tomatoes have the pH of 4.3.1657

How many times more hydrogen ions are in tomatoes than milk?1663

First, we will look at milk; and for milk, we will describe its hydrogen ion concentration with the letter M.1666

So, its pH was 6.6, and that is equal to the negative log of M.1674

We move the negative over; we get -6.6 = log(M); we can raise both sides to the 10; this cancels out,1680

and we end up getting that 10-6.6 is equal to the molarity concentration of milk; cool.1689

Next, let's do tomatoes: we will use a T to denote their hydrogen ion concentration molarity.1696

So, we were told 4.3 is equal to -log(T); -4.3 = log(T); once again, we raise them both up with a 10 underneath.1706

So, we get 10-4.3 (let me keep the same color there, so we see what is going on) is equal to...1719

and remember: since 10 and common log (log base 10) cancel out, we are left with just t.1728

So, 10-4.3 is the molarity concentration of ions in tomatoes; 10-6.6 is the molarity concentration of milk.1735

If we want to look at the ratio of ions in tomatoes to milk, the ratio of T to milk,1743

well, then we just plug in the numbers we have; we have 10-4.3 divided by 10-6.6,1754

which comes out to be 102.3, which is about 200 times.1764

So, the ratio of hydrogen ions in tomatoes to the hydrogen ions in milk is 200 times more hydrogen ions in tomatoes than there are in milk.1774

Now, let's look at lemon juice: if we have lemons--and we will use an L to denote it--1787

then we were told that 2.2 = -log(L); let me use a curly L, so we don't get confused with l in our log.1797

So, negative on both sides: -2.2 = log of our lemons.1806

We raise both sides to the 10; that cancels out with the common log,1814

so we have that 10-2.2 is equal to the molarity concentration of hydrogen ions in lemons.1819

At this point, we can now compare lemon juice to milk and tomatoes; let's compare it to tomatoes first.1827

Ratio of lemons to tomatoes is going to be 10-2.2 over tomatoes at 10-4.3,1833

which is equal to 102.1, which comes out to be around 125 times--really close; 126, actually, is closer.1844

But it is 125 times, so there are 125 times more hydrogen ions in lemons than in tomatoes.1858

Finally, let's look at the huge value that it is for milk versus lemons.1865

The ratio of hydrogen ions in lemons compared to milk is equal to 10-2.2 divided by 10-6.6,1870

which comes out to be 104.4, which is approximately 25 thousand times.1883

So, there are 25 thousand times more hydrogen ions in lemon juice than there are in milk--25 thousand!1893

This is what we are seeing with the pH scale--it is a logarithmic scale.1903

It is not just showing that there is a difference of 4.4 between milk and lemon juice.1906

We are showing this massive difference, because we are being able to encapsulate1913

this information of a huge difference with a fairly small number being able to show that.1916

We are using a logarithmic scale, because there can be these massive differences in pH reading.1921

All right, the final set of examples: Newton's Law of Cooling:1925

If you take a warm mug of tea outside on a cold winter day, the tea will cool down over time.1928

It is in this cold environment, so it is going to cool down, because it is warm, and the outside is colder.1934

So, it is going to give off its heat energy to the surrounding environment.1940

The surrounding environment is colder than it, so it will end up sucking out that energy from our mug of tea.1943

Conversely, if you put a warm mug of tea into a hot oven--we have this hot oven,1948

and we put the mug of tea in--it is going to heat up, because the hot oven is a hot environment.1953

It is hotter than the mug of tea is, so it is going to put heat energy into the mug of tea.1959

Our mug of tea will absorb heat energy from its surrounding environment.1965

If we have a hot object in a cold environment, the hot object will radiate out energy, and it will go to the cold environment.1970

If we have a cold object in a hot environment, it will pick up that energy and become hot, just like the environment it is in.1977

The rate at which heat is transferred is proportional to the difference in temperature between the object and the surrounding environment.1983

That makes a lot of sense: if you put your hand outside of a car window, and it is, say,1989

a warm day outside--a little cool--it is going to cool off your hand, slowly but surely.1998

But if you put your hand out of a car window on a freezing day, where it is colder than freezing,2003

it is going to actually cause your hand to drop in temperature really, really, really quickly.2009

So, this makes sense: if something is put into a really hot environment, it is going to heat up faster than if it is only put into a pretty hot environment.2014

The rate at which heat is transferred is proportional to the difference in temperature between the object and the surrounding environment.2020

Since the amount of heat transfer is connected to a proportion of what this difference is, it ends up drawing on exponential equations.2025

This idea is called Newton's Law of Cooling, and it is modeled by the temperature of the object2033

is equal to the temperature of the surrounding environment, added to the quantity (the object's initial temperature,2040

minus the temperature of the surrounding environment) times ekt.2046

This is the idea...the ekt is always going to end up coming out to be...k will end up being a negative number,2051

always, in these cases, when we are working these problems, because the environment2058

is going to cause this part here to eventually go and drop down to 0,2061

until eventually the object becomes the surrounding environment.2066

It will make more sense as we see some examples; so let's get started.2070

Example 4: A pot of soup is on a stove at a temperature of 85 degrees centigrade when you turn off the stove.2076

The kitchen is at 20 degrees, and after 10 minutes, the pot has cooled to 70 degrees.2082

If the hottest temperature you can eat the soup at is 55 degrees, how much longer do you need to wait?2086

All right, let's figure out our things here: what is the initial temperature of our soup?2091

The initial temperature of the soup is 85 degrees; we are told that it started at 85 degrees.2096

We are told that the kitchen is at 20 degrees; so the surrounding environment temperature is 20.2102

Now, we don't know what the ratio is; we don't know what the proportionality constant is yet.2109

And we don't know what the time is going to be in all of these cases.2113

However, we do have one specific example: at t = 10, we know that it cools to 70.2116

So, let's look at that: at time = 10 minutes, we know that the temperature of our object is 70.2122

And then, we can set up everything else: the surrounding environment is 20,2130

plus the initial temperature of 85, minus the surrounding at 20, e to the k,2133

and we know that it is 10 minutes: t = 10 in this case; so times 10--great.2139

We start working this out; we start working towards the ek(10).2145

We subtract 20 on both sides; 85 - 20 in there becomes 65, times e to the k times 10.2150

Divide by 65 on both sides; we have 50/65 = e10k.2157

Take the natural log of both sides; we have ln(50/65) = 10k, because the e disappears when I take the natural log of both sides.2162

We divide by 10 on both sides; we have ln(50/65), divided by 10, equals k.2171

And that ends up coming out to be that k is approximately equal to -0.02624.2178

So, because of the fact that this proportionality constant came out to be negative,2190

and it always will come out to be negative whenever we are looking at a Newton's Law of Cooling problem,2193

because it came out to be negative, we are going to end up seeing that this part here...2197

as t becomes larger and larger, e to the negative larger and larger thing will become smaller and smaller.2203

And this Ti - Ts portion will get crushed down to 0 as the e to the negative thing becomes smaller and smaller.2209

It will crush that down, and we will end up being left at just the temperature of our surrounding environment.2215

All right, let's figure out the other half here.2220

We want to figure out at t = [what?] we end up having a temperature of 55 degrees, because at that point we will be able to eat our soup.2222

So, at 55 temperature, we know we will have surrounding at 20, plus initial temperature was 85, minus 20, times e to our constant, -0.02624.2232

I will just leave it as k right now, and we can swap it out later when we need to figure out actual numbers.2250

That is one trick to avoid having to do a lot of writing; so ekt.2255

Subtract 20 on both sides; we get 35 = 65ekt.2259

Don't worry: we know what k is; we will just plug it in when we get to needing it.2264

35/65...we can take the natural log of both sides in just a moment to get rid of that e.2268

We have ln(35/65) = kt; divide by k on both sides; we can now plug that in, and we have ln(35/65)/-0.02624;2274

that is equal to our time, and that ends up coming out to be approximately 23.6.2293

So, 23.6 minutes from the moment that we turn off the stove (because when we turned off the stove, that was our time of T = 0)2301

is going to be when the soup is at 55 degrees Celsius.2322

However, we were asked how much longer we have to wait.2325

So, we have already waited 10 minutes, because at 10 minutes, we were told 70 degrees.2329

So, how much longer do we need to wait?2334

We need to wait 23.6 (what we got here) minus the 10 minutes that we have already waited, so 13.6 minutes, until the soup is cool enough to eat it.2336

All right, the final example--this one is going to be a little bit complicated.2359

But this is about as hard as we will end up seeing any of this sort of thing get.2364

All right, police discover a body in an air-conditioned room with a temperature of 22 degrees centigrade (Celsius).2368

When the medical examiner inspects the body at 10 PM, it has a temperature of 33 degrees.2374

At 11 PM, it is at 31 degrees; when did the person die?2379

We are also told that human body temperature is 37 degrees.2384

And finally, we are also told to give our answer to the nearest tenth of an hour.2387

All right, what is the idea going on here?2391

The police find this body, and so they can take the temperature of the body at one moment in time.2393

And then, they wait a little bit longer, and they can figure out that the body has cooled.2398

From that, they can figure out how fast the body cools.2402

And because they know what it was at one time, they can go backwards.2405

They can use the math to go backwards and figure out when the body was at a normal human body temperature--when the body was still alive.2408

And because, when it is 37 degrees, that will be the moment of death, that will tell us when the person died.2416

So, this forensic science; this is the sort of information, the sort of math,2421

that crime scene investigators can use to figure out when a person died, and be able to create a good murder case based on that.2424

All right, let's get our pieces of data from this.2432

We are told that the room has a temperature of 22 degrees.2434

So, the surrounding temperature of our room is 22; the initial temperature of our body is 37 degrees.2437

Human bodies are 37 degrees when they are alive.2446

When the medical examiner inspects the body at 10 PM, it has a temperature of 33 degrees.2449

So now, we have this confusing thing of time.2453

In the last example, we knew what time 0 meant; we knew where it was, and everything was told relative to time 0.2456

Time 0 was when you turned off the stove, and then time t = 10 was 10 minutes from turning off the stove.2462

But at this point, the t = 0 is time of death; but we don't know what 10 PM is to that yet.2468

But let's start by saying t = 0 is time of death; the moment the person dies is t = 0.2474

And from there, we are just counting there; so t = 0 is time of death; t = 1 is one hour after time of death,2486

because currently we are dealing with 10 PM, and our answer was told to be given in hours; so let's work with hours as our form here.2492

t = 0 is time of death, so t = 1 is one hour after death; t = 2 is two hours after death; t = 3 is three hours after death, etc., etc.2500

Now, that still doesn't quite tell us what 10 PM is, so we are going to have to name symbols for this.2510

Let's say that t10 is equal to the hours to get from the moment of death to 10 PM.2514

And similarly, t11 will be the hours to 11 PM.2525

Great; with all of these ideas in mind, we are ready to start working things out.2531

So, at time of 10 PM (which...we don't know how many hours it is after death, but we can still talk about it as t10),2535

we know that the body is at 33 degrees at 10 PM; we have that 33 degrees at 10 PM is equal to2544

surrounding temperature, 22, plus initial temperature, 37,2551

minus surrounding of 22ek(t10), because that is our time, t10.2556

OK, subtract 22 on both sides: we get 11 = 37 - 22 is 15ekt10.2563

So, we divide by that; we have 11/15 = ekt10.2571

Finally, we take the natural log of both sides, and we have ln(11/15) = kt10.2577

Now, at this point, we say, "Oh, no, there are two unknowns, and this is only one equation."2584

Well, we are going to have to bring a little bit more information to the table.2589

Let's look at the 11 PM hour: at t11 (that is not t = 11; that is t11,2592

which is just the name that we gave to the number of hours after the time of death when it gets to 11 PM),2600

how many hours after death is 11 PM?--at t11, we have a temperature of 31 degrees, at 11 PM.2608

And then, the surrounding environment is still 22, plus initial temperature was still 37, minus 22e to the k...2620

and now we are using a time of 11, t11.2628

Once again, I want to just point out that it is not time equals 11; it is definitely not that.2633

It is just that we named this thing, hours to 10 PM, as t10, and hours to 11 PM as t11.2638

They could be 1 and 2; they could be 5 and 6; they could be 80 and 81; we don't know yet.2644

We just came up with names for these things so that we could start working things out.2651

We can subtract 22 on both sides; we will get 9 = 37 - 22 is 15, ekt11.2654

We can divide by the 15; we get 9/15 = ekt11.2662

Take the natural log of both sides: ln(9/15) = kt11.2668

Oh, no; we have, once again, two unknowns and one equation.2674

And between all of them, we have k, t10, t11; that is 3 unknowns and 2 equations.2678

We need some other piece of information that will connect these things together.2685

How is k connected to t10? How is t10 connected to t11?2688

Of course--how is t10 connected to t11? Well, 11 PM is one hour after 10 PM!2692

To get from 10 PM to 11 PM, you go up one hour; so however many hours after death t10 is,2700

we know that, if we add one to that, we will end up being at the 11 PM mark.2706

t10 + 1 = t11; this key realization will allow us to solve things.2710

We can plug this in over here; so we have ln(9/15) = kt10 + 1.2717

kt10 + k...over here, we know what kt10 is--it is ln(11/15).2727

Let's bring this back down; ln(9/15) is equal to kt10; we swap that out for ln(11/15) + k.2735

We subtract ln(9/15) - ln(11/15) = k.2745

Now, at this point, we could just punch that into a calculator, or we could remember the properties that we know,2752

if we want to do a little bit less on our calculator.2756

That is ln(9/15), divided (because subtraction outside of logarithms becomes division inside) by 11/15.2758

So, since we are dividing by 15 on the top and the bottom, they cancel out, and we are left with ln(9/11) is equal to k.2767

We take ln(9/11) in our calculators, and we end up getting -0.20067 as our constant for k.2773

Our proportionality constant of k is -0.20067; great.2787

That means we can now take this piece of information right here.2793

We know what k is; we put that in over here; we have ln(11/15) is equal to...sorry, I accidentally wrote an extra 0...-0.20067t10.2796

We divide by that; we have natural log of 11/15, divided by -0.20067, equals t10.2814

So, t10 comes out to be exactly...well, not exactly; we will truncate it; we will cut it down,2823

and cut off some of those decimal places; we will round it to 1.5456.2835

It comes out to be approximately 1.5456.2840

However, they asked for the nearest tenth of an hour, so that ends up being about 1.5 hours.2844

Now, we want to know what time the person died at--when did the person die?2851

Well, if we are at t10, that is the hours to get to 10 PM from time of death.2856

So, if it is 1.5 hours to get to t10, then that means that death occurs 1.5 hours before t10.2862

So, we know that 10 PM is 1.5 hours after death.2870

If we go back 1.5 hours from 10 PM, we get 8:30 PM as the time of death.2882

Great; and this idea right here is actually a basis for something that real detectives2895

and real medical examiners can use--this sort of idea of how these things work out.2902

Math has a lot of really powerful applications, and this is just one example that is something that is almost out of a detective story.2906

All right, cool; we will see you at Educator.com later.2914

This is the end of logarithms and exponential things, so we are going to move into a new topic.2917

I hope everything here made sense, because it is time to get started on something else.2921

All right, goodbye!2925