For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

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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:

It willPe ^{rt}.__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 Pe^{rt}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!

- 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

where T is the object's temperature, TT = T _{s}+ (T_{i}− T_{s}) e^{kt},_{s}is the surrounding environment, T_{i}is the object's initial temperature, k is a proportionality constant, and t is the time.

### 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

### Math Analysis Online

### 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 understanding*0040

*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 decay*0066

*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 value*0109

*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 knowing*0269

*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 thing*0374

*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 problem*0426

*(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 different*0460

*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 alive*1093

*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 situation*1475

*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 same*1550

*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 encapsulate*1913

*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 object*2033

*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 environment*2058

*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 to*2544

*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 detectives*2895

*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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