### Exponential Growth and Decay

- If growth or decay is occurring by a fixed percentage during each period of time, use the formula y = a(1 + r)
^{t}or y = a(1 â€“ r)^{t}. For scientific applications, use the formula y = ae^{kt}or y = ae^{-kt}.

### Exponential Growth and Decay

depreciate at a rate of .96% per year. What is it's value after 5 years?

- Use the formula y = a(1 − r)
^{t}where a = initial amount, r is the rate and t is the number of years - y = a(1 − r)
^{t} - y = 22999.99(1 − 0.0996)
^{5} - y = 22999.99(0.9004)
^{5} - y = 22999.99(0.5918)
- y = 13611.47

depreciate at a rate of % 15.00 per year. What is it's value after 4 years?

- Use the formula y = a(1 − r)
^{t}where a = initial amount, r is the rate and t is the number of years - y = a(1 − r)
^{t} - y = 25000(1 − 0.15)
^{4} - y = 25000(0.85)
^{4} - y = 25000(0.522)
- y = 13,050.16

How long would it take for this small apartment - home to tripple in value?

- Use the equation y = a(1 + r)
^{t}. Find t when y = 3a - 3a = a(1 + 0.05)
^{t} - 3a = a(1.005)
^{t} - 3 = (1.005)
^{t} - Take the log of both sides
- log(3) = log(1.005
^{t}) - log(3) = tlog(1.005)
- t = [log(3)/log(1.005)] = 220.27

How long would it take for the price of gold to double in value?

- Use the equation y = a(1 + r)
^{t}. Find t when y = 2a - 2a = a(1 + 0.0745)
^{t} - 2a = a(1.0745)
^{t} - 2 = (1.0745)
^{t} - Take the log of both sides
- log(2) = log(1.0745
^{t}) - log(2) = tlog(1.0745)
- t = [log(2)/log(1.0745)] = 9.65

How long will it take for this investment to reach 2.5 times its original value?

- Use the equation y = a(1 + r)
^{t}. Find t when y = 2.5a - 2.5a = a(1 + 0.0805)
^{t} - 2.5a = a(1.0805)
^{t} - 2.5 = (1.0805)
^{t} - Take the log of both sides
- log(2.5) = log(1.0805
^{t}) - log(2.5) = tlog(1.0805)
- t = [log(2.5)/log(1.0805)] = 11.83

^{ − kt}, find the coefficient k.

- Set up the equation. y = 1/2a
- y = ae
^{ − kt} - [1/2]a = ae
^{ − k*5} - [1/2] = e
^{ − k*5} - Take the ln of both sides
- ln([1/2]) = lne
^{ − k*5} - ln(0.5) = − k*5

^{ − kt}, find the coefficient k.

- Set up the equation. y = 1/2a
- y = ae
^{ − kt} - [1/2]a = ae
^{ − k*0.75} - [1/2] = e
^{ − k*0.75} - Take the ln of both sides
- ln([1/2]) = lne
^{ − k*75} - ln(0.5) = − k*0.75

^{kt}. At the beginning of the experiment there

were 250,000 bacterial. Ten hours later, there were 500,000. What is the coefficient k?

- Set up the equation. y = 2a since the bacterial doubled in size.
- y = ae
^{kt} - 2a = ae
^{k*10} - 2 = e
^{k*10} - Take the ln of both sides
- ln(2) = lne
^{k*10} - ln(2) = k*10

^{kt}. At the beginning of the experiment there

were 250,000 bacterial. Ten hours later, there were 500,000. How long would it take for the

bacterial culture to reach 999,999 bacterial?

- Set up the equation. use the value of k = 0.0693 from the previous problem
- y = ae
^{kt} - 999,999 = 250,000e
^{0.0693*t} - 3.999996 = e
^{0.0693*t} - Take the ln of both sides
- ln(3.999996) = lne
^{0.0693*t} - ln(3.999996) = 0.0693*t

^{kt}. At the beginning of the experiment there

were 250,000 bacterial. Ten hours later, there were 500,000. How long did it take the bacterial to

grow to 375,000?

- Set up the equation. use the value of k = 0.0693 from the previous problem
- y = ae
^{kt} - 375,000 = 250,000e
^{0.0693*t} - 1.5 = e
^{0.0693*t} - Take the ln of both sides
- ln(1.5) = lne
^{0.0693*t} - ln(1.5) = 0.0693*t

*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

### Exponential Growth and Decay

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
- Decay 0:17
- Decreases by Fixed Percentage
- Rate of Decay
- Example: Finance
- Scientific Model of Decay 3:37
- Exponential Decay
- Radioactive Decay
- Example: Half Life
- Growth 9:06
- Increases by Fixed Percentage
- Example: Finance
- Scientific Model of Growth 11:35
- Population Growth
- Example: Growth
- Example 1: Computer Price 14:00
- Example 2: Stock Price 15:46
- Example 3: Medicine Disintegration 19:10
- Example 4: Population Growth 22:33

### Algebra 2

### Transcription: Exponential Growth and Decay

*Welcome to Educator.com.*0000

*In today's lesson, we are going to go over exponential growth and decay.*0002

*This topic was introduced briefly in a previous lecture on exponential equations.*0006

*And now, we are going to look at it in greater depth and work on some sample problems.*0011

*First, decay: there are actually two formulas that we are going to go over for decay, and two for growth.*0017

*The first decay formula talks about a situation where you have a quantity that decreases by a fixed percentage each year.*0024

*The equation for this situation is y = a(1 - r) ^{t}.*0033

*This gives the amount, y, of the quantity after t years; so a is the initial amount that you begin with;*0040

*r is the percent decrease; and in order for this formula to work correctly, you need to express it as a decimal.*0056

*y is the amount that you end up with, the quantity after t time; and t is the time that has passed.*0072

*A typical application for this formula would be in finance.*0091

*For example, if you have $1,000 in your bank account, and you remove 40% of whatever you have in that account each year,*0096

*and you are trying to figure out how much you would have left in 3 years, you could use this formula.*0105

*So, I said you began with $1,000; therefore, a = 1,000 dollars--that is how much money you started out with in your savings account.*0111

*The percent decrease is going to be 40%, because you are taking that amount out each year (or .4).*0122

*And we are trying to figure out how much money you are going to have left after 3 years,*0132

*so the unknown is the amount of money that you have left after 3 years; and the time is 3 years.*0138

*So, trying this problem out using this formula: we are going to take y =...and y is our unknown;*0146

*a is 1,000; times 1 minus...and it is .4, expressed as a decimal, raised to the t power (which here is 3).*0155

*That gives me y = 1000(0.6) ^{3}; .6^{2} is .36, times .6 is .216; so this leaves me with y = 1000(.216).*0166

*And multiplying that out, I would find that I would have 216 dollars left in my account after 3 years of removing 40% of the money in the account.*0184

*So again, this decay formula is the general formula often used for applications, talking about financial applications, interest...*0196

*Actually, interest would be growth...but interest payments...decreases; we are talking about decreases.*0207

*The scientific model of decay is slightly different.*0217

*And in a lot of applications in science, the formula that is used is exponential decay: y = ae ^{-kt}.*0221

*We talked previously about how base e is often used in scientific applications; we see it popping up here again.*0231

*And as in the last formula, y is the amount that you end up with after t years, whereas a is the initial amount.*0237

*This time, instead of having a fixed percent that it decreases by, we use a constant.*0245

*And the constant depends on the situation.*0251

*A typical situation would involve radioactive decay; and with radioactive decay, or any type of decay, you will be given a constant.*0253

*There might be a certain isotope or a certain compound that decays according to one k.*0262

*And then, you look at a different compound, and it might decay a lot faster; and then the value of k would be different.*0269

*So, just to review what the graph would look like when we are talking about an exponential decrease:*0274

*the shape of the graph would be roughly like this, over time.*0279

*You see here: as time moves forward, the amount that you have left is getting smaller and smaller and smaller and smaller, and approaching 0 eventually.*0284

*This formula, as I mentioned, is important when working with radioactive substances and thinking about radioactive decay.*0298

*And you may have heard of the concept of half-life.*0305

*Half-life is the amount of time it takes for half of the material to decay away.*0308

*So, let's say I have 10 grams of a radioactive substance, and it takes 3 months for it to decay until there are only 5 grams left.*0313

*The half-life of that substance would be 3 months, the amount of time it takes for half of it to be gone.*0326

*And let's look at an example of this: let's say that I have a radioactive compound that decays exponentially according to this formula.*0333

*And k equals 0.1 for a certain radioactive substance; and it follows exponential decay; and I want to know the half-life.*0341

*All right, let's look at this formula and think about how this will work.*0361

*I have y = ae ^{-kt}, and I am not given the original amount, which seems like it would be a problem;*0364

*but it is actually not, and here is why: think about what half-life is saying.*0370

*It is saying that half of the original amount is left.*0375

*So here, the amount we are left with, y, equals half the original amount, or 1/2a.*0382

*Knowing that, I can set this up as follows: 1/2a = ae ^{-kt}.*0390

*Once I have this set up like this, the a's will cancel; I am going to divide both sides (I am trying to get rid of the a on the right) by a.*0397

*That cancels; this cancels; and I just am left with 1/2 = e ^{-kt}.*0406

*Recall, earlier on, when we were working with exponential equations, we talked about how it is possible to solve these by taking the log of both sides.*0414

*So, let's go ahead and fill in what we have for k, which is -.1.*0426

*And what we are looking for is the time--the amount of time it takes for half of the substance to decay.*0432

*So, we have it down to this point; and what I can do is take a log of both sides.*0440

*And since I am working with base e, I am going to go ahead and work with the natural log,*0444

*although you actually could use any log, as long as it has the same base.*0448

*That is going to give me ln(1/2) = ln(e ^{-.1t}).*0451

*Because these are inverses, the ln(e)...these are essentially going to cancel each other out.*0461

*And on the right, I am going to be left with -.1t.*0468

*On the left, I am going to actually rewrite this as ln(2 ^{-1}); and recall that 2^{-1} is the same as 1/2.*0475

*All right, let's go on up here to finish this one out.*0484

*ln(2 ^{-1}) = -.1t: recall, from the rules of working with logs, that I can rewrite this as -1ln(2), or just -ln(2), equals -.1t.*0486

*I am going to divide both sides by -.1; the negatives are going to cancel out, and this is going to give me ln(2)/.1 = t.*0508

*So, I am rewriting this like this, ln(2) divided by .1.*0517

*You can use your calculator and the natural log button on there to find out what ln(2) is, divided by .1.*0526

*And that actually comes out to approximately 7 months; so it would take 7 months for this compound to decay until there is only half left.*0533

*Therefore, the half-life is 7 months.*0541

*That was decay; now we are talking about growth.*0546

*OK, so again, there are two formulas, and they are very similar to the decay formulas.*0554

*Here, we are talking about growth, so for a quantity that increases by a fixed percentage each year, the equation...*0558

*this actually should be a plus right there...y = a(1 + r) ^{t} gives the amount, y, of the quantity after t years.*0571

*So, a is the initial amount, just like we talked about with decay.*0583

*And r is the percent increase, and we need to express that as a decimal.*0588

*And what we are looking for is the final amount, the amount that we are going to end up with.*0592

*And this is the general formula; so this is the model that you might see used again when working with money or financial-type problems.*0597

*Let's look at an example: let's say that you had $500 in your bank account, so a = 500.*0607

*And you receive 10% interest per year on that.*0616

*I am going to rewrite that in a decimal form: .1.*0620

*And what you want to know is how much money you are going to end up with after 3 years.*0624

*So, rewriting this as y = a(1 + r) ^{t}, this is going to give me y = 500 (that is what I started out with);*0632

*and this is 1 + .1, raised to the t power; and in this case, t is actually 3--let's go ahead and make that 3.*0646

*y = 500(1.1) ^{3}*0662

*And you could use your calculator to determine that 1.1 raised to the third power is equal to 1.331.*0669

*Multiplying that by 500, you would end up with 665 dollars and 50 cents in your account after 3 years, if you are receiving 10% interest on that $500.*0678

*So, you can see that this is a really useful formula.*0690

*The scientific model of growth is analogous to the scientific model of decay that we talked about before.*0695

*Notice here that now we have a positive exponent; with the decay formula, y = ae...it was -kt; now we have a positive exponent.*0700

*And the constant k depends on the situation.*0711

*You might be asked to find it, if they give you what y is, and a, and t; or you may be given the k and then asked to figure out something else.*0715

*An important application here in science is in population growth.*0724

*And systems that follow this model of growth increase exponentially, so their graph is going to look approximately like this.*0728

*So, let's consider an example where you have a population of 1,000 people.*0740

*There are 1,000 people living in a town, and this town's population is increasing according to this exponential model.*0745

*And k is 0.2: we want to know the number of people that we are going to end up with in 10 years.*0756

*So, using this formula, y = ae ^{kt}, we don't know y; we know that a is 1,000; e; k is .2; and time is 10 years.*0770

*This gives me y = 1,000e...0.2 times 10--that is just going to give me 2.*0790

*Now, I can figure this out; I could use my calculator, or you might remember that e is 2.7182.*0797

*So, if I square that, I will get approximately (for e ^{2}) 7.389.*0805

*Multiply by 1,000: I am going to end up with 7,389 people living in the town 10 years down the road.*0817

*OK, so that was the four equations that we are working with today: two each for growth and decay.*0831

*So, let's try some examples: A computer is purchased for $2,000, and it is expected to depreciate at the rate of 20% per year.*0840

*Depreciate means it is losing value: so you bought your computer, and each year it is worth less,*0851

*which is frequently the case with objects that you buy.*0858

*Let's think about what formula we are going to use.*0862

*Since this is decreasing, we know that we need to use a decay formula.*0864

*And since it is decreasing at a certain percentage per year, I am going to use the general decay formula, which is y = a(1 - r) ^{t}.*0868

*My initial amount is $2,000; that is what I purchased it for.*0880

*And the rate of depreciation is 20%; in decimal form, this is .2.*0884

*I want to know its value after a certain amount of time; and the amount of time is 3 years.*0892

*With this in mind, I can just go ahead and work this out.*0901

*y is what I am looking for; a is 2,000; times 1 - .2, raised to the third power.*0906

*Therefore, y = 2,000...1 minus .2 is .8, to the third power; you can go ahead and work this out on your calculator;*0914

*and you will find that it comes out to .512, times 2,000, is 1,024 dollars.*0926

*So, this computer, in three years, at this rate of depreciation, is going to be worth $1,024.*0935

*Now, we are talking about a stock that is expected to increase in value.*0947

*Since we are talking about increasing, we are talking about growth.*0950

*And this is at the rate of 45% per year.*0953

*Since I am talking about a steady increase that is based on percent per year, I am going to use the general growth formula, not the scientific exponential one.*0958

*So, I am going to use this formula: y = a(1 + r) ^{t}.*0968

*And I want to know how long it will take for the stock to triple in value.*0975

*So, you have to think carefully about how to set this up.*0979

*I know the rate: the rate is given as 45%, which is equal to .45.*0982

*And I am asked how long it will take for the stock to triple in value; so I want to know*0992

*when the value I am going to end up with, y, is 3 times the initial value, 3a.*0997

*So, I have an initial value, a: it doesn't matter what that value is.*1003

*What I want to know is when it is going to be 3 times whatever a is.*1006

*So, if you just keep going on this, you see that the a will drop out.*1010

*So then, I set this up as...let's go ahead and start it over here...for y, I am going to substitute 3a.*1013

*This equals a; I don't know what a is--I just put a.*1021

*1 + .45...and I know that what I am looking for, actually, is t: so, I just leave this as t.*1024

*When I divide, if I want to move this a over here, I am going to divide both sides by a; and the a's will conveniently drop out,*1036

*leaving us with 1 + .45, raised to the t power: so 3 = 1.45 ^{t}.*1045

*Recall that, when you are working with exponential equations with different bases*1053

*that you can't easily get to become the same base, you can just take the log of both sides.*1057

*And you can use any base log you want; but two convenient ones are the natural log and the common log.*1063

*So, I am going to go ahead and take the common log, the base 10 log, of both sides.*1070

*log(3) = log(1.45 ^{t}).*1075

*Using properties of logs, I can rewrite this as t times log(1.45); and I am looking for t.*1081

*I want to isolate t, so I am going to make this log(3), divided by log(1.45), dividing both sides by log(1.45).*1092

*And remember that these are just numbers; they have a value--I am going to work with them and move them around, just like I would any other numbers.*1103

*I am rewriting this so that the t is on the left--a more common form.*1112

*This is something you can then use your calculator to figure out.*1118

*And if you take the log of 3, divided by log base 10 of 1.45, you will get approximately equal to 3 years.*1121

*So, even though I didn't know what my stock was worth to start with, it didn't matter.*1134

*All I wanted to know is how long it is going to take whatever amount I have to triple; and it is going to take 3 years.*1139

*Example 3: A medicine disintegrates in the body at a steady rate.*1153

*We are talking about disintegration, which is a type of decrease; so we are talking about decay.*1158

*It decays based on the equation y = ae ^{-kt}.*1163

*This is using the scientific model of decay with base e, where k equals 0.125, and t is in hours.*1168

*Find the half-life of this medicine.*1178

*We are using this formula; let's rewrite it here; and we want to find the half-life.*1182

*So, we are looking for the half-life, which is often written this way: t ^{1/2}.*1186

*And I don't know how much medicine we are starting out with, but it doesn't matter.*1195

*It is similar to the last one we worked out, where it doesn't matter what you are starting out with.*1201

*You are just looking for a change.*1204

*I know k; that is given; and I know that what I am going to end up with is half of the initial.*1208

*Since I am looking for half-life, after this amount of time, y is going to be equal to half of the amount that we started out with, 1/2a.*1218

*I am coming over here and setting this up, substituting 1/2a for y, equals ae, and then I have -.125 as k; and what I am looking for is t.*1226

*If I divide both sides by a, the a will drop out.*1243

*This gives me 1/2 = e ^{-.125t}.*1246

*One way to solve this is to take the log of both sides; I am going to go ahead and take the natural log of both sides; this is ln(1/2) = ln(e ^{-.125t}).*1255

*Because these are inverses, I am going to end up with -.125t.*1271

*To go further with this, I am going to rewrite this left part, ln(1/2), as ln of 2 raised to the -1 power,*1277

*just like the example we worked out a few slides ago, equals -.125 times t.*1286

*I can bring this out in front, which would be -1ln(2), or just -ln(2), equals -.125t.*1298

*I am going to divide both sides by -.125, and this equals t; so these negatives...a negative divided by a negative*1307

*is going to become a positive, so I come up here and get ln(2) divided by .125 = t.*1320

*Now, I can find the natural log of 2 and divide that by .125, using a calculator.*1329

*And it turns out that this is approximately equal to 5 hours.*1335

*Therefore, the half-life of this medication, given this constant, is approximately 5 hours, based on using this model for exponential decay.*1340

*Example 4: The population growth of a city can be modeled exponentially with a constant of k = 0.01.*1355

*The current population is 100,000; what will it be in 100 years?*1363

*Since this is exponential growth, and we are given a constant, I am going to use the scientific model for growth, that formula, y = ae ^{kt}.*1368

*All right, I have been given a k; this is 0.01; and I have been given the original population, which is 100,000.*1382

*I have been given a t of 100 years, and I am asked to find y.*1395

*I want to know what the population is going to end up being after 100 years.*1401

*Therefore, this becomes y = 100,000 times e; and that would be a k of .01, times a t of 100.*1407

*Therefore, this is y = 100,000 times e; and this is just going to be .01 times 100, so that is going to give me 1.*1418

*Recall that e is 2.71828, so instead of e to the first power, it is just e, which I know the value of.*1428

*Then, I go ahead and multiply this times 100,000 to get that 271,828 will be the population in 100 years.*1441

*Again, this was simply a problem involving the use of the scientific model for growth.*1456

*That concludes this lesson on Educator.com on exponential growth and decay; thanks for visiting!*1464

1 answer

Last reply by: Dr Carleen Eaton

Mon May 4, 2015 9:10 PM

Post by David Saver on May 1, 2015

In the Half-life example, where does the answer 7 months come from?

Or is it suppose to be 7 years?

1 answer

Last reply by: Dr Carleen Eaton

Mon Nov 25, 2013 9:41 PM

Post by Mirza Baig on November 25, 2013

What the difference between this formula y=a*p^a and y=a(1+r)^t????

Can we use any two formula given below and find the growth of change ??

1 answer

Last reply by: Dr Carleen Eaton

Mon May 4, 2015 9:10 PM

Post by Adela Bravo on May 30, 2012

At 8:27 shouldn't 7months be 7years, because t=years.

1 answer

Last reply by: Dr Carleen Eaton

Sun Nov 27, 2011 4:11 PM

Post by Vanza Bik on November 26, 2011

I have a question? How can I ask you?