For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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:
Pe^{rt}. - 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!
- 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 = T_{s} + (T_{i} − T_{s}) e^{kt},
Application of Exponential and Logarithmic Functions
- 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):
where P is the starting quantity, r is the rate coefficient, and t is the time elapsed.A = Pe^{rt}, - 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 = 15000e^{r·10} - Solve for r:
Plugging in to a calculator, we find that r = 0.053.e^{10r} = 25484 15000⇒ 10r = ln ⎛
⎝25484 15000⎞
⎠⇒ r = ln ⎛
⎝25484 15000⎞
⎠10 - Now that we know r, we have enough information to see what the balance is at t=20:
A = 15000 e^{0.053 ·20} = 43 295.56
- 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
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.A = P ⎛
⎝1 + r n⎞
⎠nt
. - Setting it up at t=7, we have:
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.):28273 = 22000 ⎛
⎝1 + r 4⎞
⎠4·7
Plugging in to a calculator, we find r = 0.036⎛
⎝1 + r 4⎞
⎠28
= 28273 22000⇒ ⎛
⎝1 + r 4⎞
⎠= 28 ⎛
√28273 22000⇒ r = ⎛
⎜
⎝28 ⎛
√28273 22000−1 ⎞
⎟
⎠·4 - 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:
We can do this here as well (see the lesson for an explanation of why we are able to do this).A = Pe^{rt} - Setting up with this formula, we can solve for r using information about the balance at t=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 Pe^{rt} instead of the standard formula, you will find a different value for your r rate.]28273 = 22000 e^{r·7} ⇒ r·7 = ln ⎛
⎝28273 22000⎞
⎠⇒ r = ln ⎛
⎝28273 22000⎞
⎠7 - Now we can find the balance at t=15:
This goes to show that in many situations where exponential growth is involved, we can often just default to using Pe^{rt} if we so desire.A = 22000 e^{0.035838 ·15} ≈ 37 661
- 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):
where P is the starting quantity, r is the rate coefficient, and t is the time elapsed.A = Pe^{rt}, - For this problem, we are looking to solve for r. Set up the equation:
19400 = 7000e^{r·20} - Solve for r:
Plugging into a calculator, we find r = 0.050 968.e^{20r} = 19400 7000⇒ 20 r = ln ⎛
⎝19400 7000⎞
⎠⇒ r = ln ⎛
⎝19400 7000⎞
⎠20 - 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:
Finally, the problem asked for the percentage to three decimal places, so we must round to three decimal places: 5.097%.r = 0.050 968 ⇒ 5.0968%
- 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):
where P is the starting quantity, r is the rate coefficient, and t is the time elapsed.A = Pe^{rt}, - 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:
From here, we can work to solve through substitution.1858 = P e^{r·10} 3456 = Pe^{r·20} - Working with the first of the above equations, we have
We can now plug this in to the other equation:P = 1858 e^{10r}. ⎛
⎝1858 e^{10r}⎞
⎠e^{20r} = 3456 - From here, we work to solve for r:
Plugging into a calculator, we find r=0.062 061.⎛
⎝1858 e^{10r}⎞
⎠e^{20r} = 3456 ⇒ 1858 e^{10r} = 3456 ⇒ 10r = ln ⎛
⎝3456 1858⎞
⎠⇒ r = ln ⎛
⎝3456 1858⎞
⎠10 - Now that we know r, we can plug it into our previous equation to find the value of P:
P = 1858 e^{10·0.062061}≈ 999
- 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 = Pe^{rt}. - 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 e^{r ·5} - Solve for r:
Using a calculator, we get r=0.884 730.e^{5r} = 417000 5000⇒ 5r = ln ⎛
⎝417 5⎞
⎠⇒ r = ln ⎛
⎝417 5⎞
⎠5 - Finally, we want to know the population at 5, which is nine hours after beginning, so t=9:
We were told to round to the nearest thousand, so we get a population of 14 357 000.A = 5000 e^{0.884730 ·9} = 14 357 215
- 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
where t is the number of years since we started with the mass of Pu-238. [We could also model this with Pe^{rt} ⇒ 1000 e^{rt}, 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 = e^{r·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 Pe^{rt} 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).]A = 1000 ⎛
⎝1 2⎞
⎠[t/87.7]
, - We want to know what t is when the amount is A=1, so we plug in and solve:
[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.]1000 ⎛
⎝1 2⎞
⎠[t/87.7]
= 1 ⇒ ⎛
⎝1 2⎞
⎠[t/87.7]
= 1 1000⇒ t 87.7log 1 2= log 1 1000
Using a calculator, we find t = 874.t 87.7= log 1 1000log 1 2⇒ t = ⎛
⎜
⎜
⎜
⎝log 1 1000log 1 2⎞
⎟
⎟
⎟
⎠·87.7
- 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
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.A = 1.000 000 · ⎛
⎝1 2⎞
⎠[t/n]
= ⎛
⎝1 2⎞
⎠[t/n]
, - Thus, we can solve for n by plugging in the information for a year later:
Plugging into a calculator, we find n=170.0.995 931 = ⎛
⎝1 2⎞
⎠[1/n]
⇒ log(0.995931) = 1 nlog 1 2⇒ n = log 1 2log0.995931 - Alternatively, we could approach this problem using the Pe^{rt} formula. First, we need to find the value of r:
Using a calculator, we find r = −0.004 077. Next, we need to figure the length of time for one half-life.0.995931 = 1 ·e^{r·1} ⇒ 0.995931 = e^{r} ⇒ ln0.995931 = r - 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
We can now solve this equation for t to find the half-life:0.5 = 1·e^{−0.004077 ·t}.
Plugging into a calculator, we get our half-life time of t=170.e^{−0.004077 t} = 0.5 ⇒ −0.004077 t = ln0.5 ⇒ t = ln0.5 −0.004077
- 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:
Plugging into our calculator, we find t ≈ 47 000.⎛
⎝1 2⎞
⎠[t/5730]
= 0.34 100⇒ t 5730log 1 2= log 0.34 100⇒ t = log 0.34 100log 1 2·5730 - As a side note, it would be possible to do this problem using Pe^{rt}, 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.
- 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
where T is the object's temperature, T_{s} is the surrounding environment, T_{i} is the object's initial temperature, k is a proportionality constant we figure out for each problem, and t is the time.T = T_{s} + (T_{i} − T_{s}) e^{kt}, - The initial temperature of the soup is T_{i} = 85 and the surround temperature in the refrigerator is T_{s}=3. At t=5, we have the equation
which we can now solve for the coefficient k.72 = 3 + (85−3) e^{k·5}, - Solving for k, we get
Using a calculator, we find k = −0.034 523.3 + 82 e^{5k} = 72 ⇒ e^{5}k = 69 82⇒ 5k = ln 69 82⇒ k = ln 69 825 - 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:
Plugging into a calculator, we find that the soup is cool enough at t ≈ 13.3 + 82e^{−0.034523t} = 55 ⇒ e^{−0.034523t} = 52 82⇒ t = ln 52 82−0.034523 - 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.
- 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
where T is the object's temperature, T_{s} is the surrounding environment, T_{i} is the object's initial temperature, k is a proportionality constant we figure out for each problem, and t is the time.T = T_{s} + (T_{i} − T_{s}) e^{kt}, - (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 t_{0}. Thus, the police measure the body's temperature at t = t_{0}, and therefore the detective measures the body's temperature 19 minutes later at t = t_{0} + 19.
- Immediately at death, the body has an initial temperature of T_{i} = 37 and the surrounding environment is T_{s} = −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:
Moving things around a bit in the equations, we get31.5 = −15 + (37 − (−15)) e^{k t0} 27.3 = −15 + (37 − (−15)) e^{k (t0+19)} 46.5 = 52e^{k t0} 42.3 = 52 e^{kt0 + 19k} - At this point, we have two unknowns: t_{0} 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
At this point, we can substitute in our first equation, 46.5 = 52e^{k t0}:42.3 = 52 e^{kt0 + 19k} ⇒ 42.3 = 52 ( e^{kt0} ·e^{19k} ) ⇒ 42.3 = 52 e^{k t0} ·e^{19k} 42.3 = 52 e^{k t0} ·e^{19k} ⇒ 42.3 = (46.5) ·e^{19k} - From here, we can now solve for k:
Using a calculator, we find k = −0.004 982.46.5 e^{19k} = 42.3 ⇒ e^{19k} = 42.3 46.5⇒ 19k = ln 42.3 46.5⇒ k = ln 42.3 46.519 - We can now plug this in to the first of our two temperature equations to solve for t_{0}:
Using a calculator, we find t_{0} = 22.439.46.5 = 52e^{−0.004982 t0} ⇒ −0.004982 t_{0} = ln 46.5 52⇒ t_{0} = ln 46.5 52−0.004982 - The problem asked for the time of death to the nearest minute, so we round to t_{0} ≈ 22. Remember, t_{0} represents the number of minutes after the victim's death. We know that t_{0} 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 T_{i} = 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) ) e^{k·19} - Solve for k:
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.)46.5 e^{19k} = 42.3 ⇒ e^{19k} = 42.3 46.5⇒ 19k = ln 42.3 46.5⇒ k = ln 42.3 46.519 - 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:
Solving for t, we find37 = −15 + (31.5 −(−15)) e^{−0.004982 t}
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 10:29.46.5 e^{−0.004982 t} = 52 ⇒ −0.004982 t = ln 52 46.5⇒ t = ln 52 46.5−0.004982
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.
*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.
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
Precalculus with Limits Online Course
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
Now, I have a secret for you: don't let anybody else know about this.0138
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 Pe^{rt} 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 Pe^{rt} formula.0236
But if you can figure out what that r has to be for Pe^{rt} 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 Pe^{rt} 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 Pe^{rt} when we have all of these special formulas?0287
How can you swap out some exponential formula for just natural exponential growth, the Pe^{rt} form?0291
The reason why is because of this e^{r} part; e^{r} 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 Pe^{rt} 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 Pe^{rt} 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 e^{r} 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 Pe^{rt} instead.0446
By choosing the right r for Pe^{rt}, 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 Pe^{rt}, 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 Pe^{rt} 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 10^{0},0550
and 10 billion is the same thing as 10^{10},because we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 zeroes,0555
so that is the same thing as 10^{0} to 10^{10}.0568
And because it is log base 10--it is the common logarithm--when log base 10 operates on 10^{0}, we will get 0.0571
When it operates on 10^{10}, 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 20^{th}, 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 radioactive: it radiates something away.0957
If you want to learn more about this, you can study either chemistry or physics to learn more about radiation.0960
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
So, let's just start with the general formula for half-life.1159
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
So, what ratio do we want to start with?1243
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 Pe^{rt} form.1399
So, the amount equals Pe^{rt}; 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 = e^{rt}.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 = e^{r(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 10^{2.3}, 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 10^{2.1}, 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 10^{4.4}, 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 e^{kt}.2046
This is the idea...the e^{kt} 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 e^{k(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 = e^{10k}.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 T_{i} - T_{s} 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 e^{kt}.2255
Subtract 20 on both sides; we get 35 = 65e^{kt}.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
And there is our answer.2357
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 t_{10} is equal to the hours to get from the moment of death to 10 PM.2514
And similarly, t_{11} 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 t_{10}),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 22e^{k(t10)}, because that is our time, t_{10}.2556
OK, subtract 22 on both sides: we get 11 = 37 - 22 is 15e^{kt10}.2563
So, we divide by that; we have 11/15 = e^{kt10}.2571
Finally, we take the natural log of both sides, and we have ln(11/15) = kt_{10}.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 t_{11} (that is not t = 11; that is t_{11},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 t_{11}, 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, t_{11}.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 t_{10}, and hours to 11 PM as t_{11}.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, e^{kt11}.2654
We can divide by the 15; we get 9/15 = e^{kt11}.2662
Take the natural log of both sides: ln(9/15) = kt_{11}.2668
Oh, no; we have, once again, two unknowns and one equation.2674
And between all of them, we have k, t_{10}, t_{11}; 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 t_{10}? How is t_{10} connected to t_{11}?2688
Of course--how is t_{10} connected to t_{11}? 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 t_{10} is,2700
we know that, if we add one to that, we will end up being at the 11 PM mark.2706
t_{10} + 1 = t_{11}; this key realization will allow us to solve things.2710
We can plug this in over here; so we have ln(9/15) = kt_{10} + 1.2717
kt_{10} + k...over here, we know what kt_{10} is--it is ln(11/15).2727
Let's bring this back down; ln(9/15) is equal to kt_{10}; 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...-0.20067t_{10}.2796
We divide by that; we have natural log of 11/15, divided by -0.20067, equals t_{10}.2814
So, t_{10} 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 t_{10}, that is the hours to get to 10 PM from time of death.2856
So, if it is 1.5 hours to get to t_{10}, then that means that death occurs 1.5 hours before t_{10}.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
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