Sign In | Subscribe
INSTRUCTORSCarleen EatonGrant Fraser
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 2
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (8)

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?

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 – r)t. For scientific applications, use the formula y = aekt or y = ae-kt.

Exponential Growth and Decay

A recent college grade bought a brand new car for $ 22,999.99. The value of the car is expected to
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
After 5 years the car will be valued at $ 13,611.47.
A high school student bought a car for for $ 25,000. The value of the car is expected to
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
After 4 years the car will be valued at $ 13,050.16
A small apartment - home is expected to increase in value at the rate of 5% per year.
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(1.005t)
  • log(3) = tlog(1.005)
  • t = [log(3)/log(1.005)] = 220.27
It will take 220.27 years for the home to tripple in value.
The value of gold is expected to increase in value at a rate of 7.45% per year.
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(1.0745t)
  • log(2) = tlog(1.0745)
  • t = [log(2)/log(1.0745)] = 9.65
It will take 9.65 years for the price of gold to double in value.
A new investment strategy promises potential investors a return of investment of 8.05% per year.
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(1.0805t)
  • log(2.5) = tlog(1.0805)
  • t = [log(2.5)/log(1.0805)] = 11.83
It will take 11.83 years for the investment to reach 2.5 its original value.
Caffeine has a half - life of 5 hours. If caffeine decays based on the equation y = ae − 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
k = [ln(0.5)/( − 5)] = 0.139
Penicillin has a half - life of 45 minutes. If penicillin decays based on the equation y = ae − 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
k = [ln(0.5)/( − 0.75)] = 0.9242
The bacterial growth in the lab can be modeled by y = aekt. 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 = aekt
  • 2a = aek*10
  • 2 = ek*10
  • Take the ln of both sides
  • ln(2) = lnek*10
  • ln(2) = k*10
k = [ln(2)/10] = 0.0693
The bacterial growth in the lab can be modeled by y = aekt. 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 = aekt
  • 999,999 = 250,000e0.0693*t
  • 3.999996 = e0.0693*t
  • Take the ln of both sides
  • ln(3.999996) = lne0.0693*t
  • ln(3.999996) = 0.0693*t
t = [ln(3.999996)/0.0693] = 20.004 hours
The bacterial growth in the lab can be modeled by y = aekt. 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 = aekt
  • 375,000 = 250,000e0.0693*t
  • 1.5 = e0.0693*t
  • Take the ln of both sides
  • ln(1.5) = lne0.0693*t
  • ln(1.5) = 0.0693*t
t = [ln(1.5)/0.0693] = 5.85 hours

*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

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; .62 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)30662

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 = aekt, 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 e2) 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 know0992

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 = 1.45t.1045

Recall that, when you are working with exponential equations with different bases1053

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(1.45t).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: t1/2.1186

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 negative1307

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 = aekt.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