Applications of Logarithms: Growth and Decay

Objectives:  To learn how logarithms are used to solve growth and decay problems.

Related topics: 1, 3-5, 9, 10, 13, 25, 30.

Recall 1:  Consider the following problem:

A population of 1000 individuals is expected to grow exponentially over the next ten years at 4% per year.  How large will the population be at the end of 10 years?

This is the sort of problem that requires an understanding of exponentials.  You would solve it by noting that the population starts with 1000 individuals and grows to
1000 + 0.04(1000) = 1.04(1000) by the end of the first year.
To find the population at the end of the second year you would add 4% of the population at the beginning of the second year to the population that was already there at the beginning of the second year.  So you would have
1.04(1000) + 0.04(1.04(1000)) = (1.04)^2(1000) individuals at the end of the second year.
You can see that at the end of the n-th year the populations would be (1.04)^n(1000).

Recall 2  You can see that if the starting population was 2000 individuals, the population at the end of n years would be (1.04)^n(2000).  And if the starting population was P0, the population at the end of n years would be (1.04)^nP0.

Recall 3 You can see that if the starting population was P0 and the rate of growth was 5%, instead of 4%, the population at the end of n years would be (1.05)^nP0.
Likewise, if the rate of growth was r% and the starting population was P0, then the population at the end of n years would be

(1+r/100)^n*P0.

Recall 4 Back to the question that was posed in Recall 1.
The answer is (1.04)^10(1000) = 1480.

Recall 5: We pose a similar question.

How many years will it take for the population to reach 2000?

To answer this question we must let the number of years be a variable, say n, and solve the equation (1.04)^n(1000) = 2000 for n.   Divide both sides of this equation by 1000.  We get (1.04)^n = 2.

Recall 6: Recall that if a^x = y, then x = logay.  Therefore, the answer to the last question is

n = log1.042.

Recall 7:  How will you calculate log1.042 if you have a calculator that only calculates ln or log?

Recall that logbc = logac/logab.

Then let a = e, b = 1.04 and c = 2.  You would have log1.042 = ln 2/ln 1.04.

Introduction 1 (Decay):

The questions that were answered above are classified as growth questions.  For example, "how long will it take for a population that increases at 4% per year grow to x individuals?" is a growth question.  Note that the word population can be replaced by any other word that indicates some collection of things; i.e., money, lilies in a pond, raindrops, etc.

The answer to the question in Recall 1 was (1.04)^10(1000) = 1480.
But if the question was about a population that shrinks at a rate of 4% per year then the answer would have been (1-.04)^(10)(1000) = 665.  Note that a negative exponent models shrinking or decaying.

Consider the following question:

A certain radioactive substance has a half-life of 40 minutes.  What fraction of an initial amount of this substance will remain after 1 hour and 20 minutes?

To answer it, you would first have to know what half-life means.  It simply means the amount of time it takes before the population reduces to half of the original amount.

The model for decay is

P(n) = P(0)e^(kn)

where P(n) represents the amount of substance (or population of atoms) after n years, P(0) represents the initial amount, and k is some constant that depends on what substance we are talking about.  If k is negative, the population will decay; if it is positive, it will grow.  The trick is to first find k.
We are given that half of the substance will remain when n = 40 minutes.  So

Therefore, 40k = ln (1/2) and

k=(-ln2)/40.

This means that we know the population at n years to be P(n) = P(0)e^{[(-ln2)/40)]n}.

Notice that we could also write this last function as

P(n) = P(0)2^(-n/40), or

P(n) = P(0)(1/2)^(n/40).

If you change the half-life from 40 to 50 minutes, you will find that the population at n years will be P(n) = P(0)2^(-n/50).  And if the half-life is h minutes, the population at n years will be

P(n) = P(0)2^(-n/h).

We may now answer the original question.  It asks for the value of P(80).  From our model, we know that P(80)/P(0) = 2^(-80/40).  This is the same fraction as 2^(-2), or 1/4.

Keep the Form in Mind

1. You may have noticed that the form of the function that gives the population is

P(n) = P(0) 2^(-n/h)

where h is the half life.  The 2 is conspicuously suggesting half-life or doubling time.

If a population doubles every d years, then the population at the end of n years will be

P(n) = P(0) 2^(n/d)

where d represents the number of years it takes for the population to double from its original size.

2.  You might guess that if h represents the third-life (the time it takes for the population to shrink to one third its size) then the population at the end of n years would be

P(n) = P(0) 3^(-n/h).

3.  You may see population functions displayed as P(n) = P(0) (1/2)^(n/h).  This is just another way of expressing P(n) = P(0) 2^(-n/h) and therefore just another way of expressing population using the half-life h.

Examples

Example 1.   You put \$1,000 in a money market fund.  It takes 10 years for your money to double.  Approximately how long will it take for your money to triple?

Solution:  P(n) = P(0)2^(n/10) .

Let n be such that P(n) = 3P(0).  Then 3 = 2^(n/10).  Therefore,

log23 = n/10 and n = 10 log23 = 15.85.

So it will take about 16 years for your money to triple.

Example 2.   A certain radioactive substance has a half-life of 40 minutes.  What percentage of an initial amount of this substance will remain after 2 hours and 40 minutes?

Solution:  Set P(n) = P(0)2^(-n/40).
The question asks you to find P(160)/P(0).  But that may be found from the equation

P(160)/P(0) = 2^(-160/40) = 2^(-4) = 1/16 = 6.25%.

Example 3.   You put \$1,000 in a money market fund at r percent interest compounded annually.  After 5 years you have \$2,000 in your account.  What was the interest rate r?

Solution:  You know that 2000 = (1+r/100)^5(1000) and hence the equation we want to solve is 2 = (1+r/100)^5 .
But 2^(1/5) = 1 + r/100 becomes

100(2^(1/5 )- 1) = r

r =14.87

Example 4.   A certain radioactive element has a half-life of 900 years.  Starting with 30 milligrams there will be q(t) milligrams left after t years, where q(t) = 30(1/2)^(kt). How much will be left after 2500 years?

Solution: We know that

q(t) = 30(1/2)^(kt)

And therefore,  15 = 30(1/2)^(k*900)   Solve this for k to get 1/2 = (1/2)^(k*900) ,
or k = 1/900.
Hence, q(2500) = 30(1/2)^[(1/900)(2500)] = 30(1/2)^(25/9) = 4.37

Example 5.   Charcoal remains, found in an archeological site, was found to contain only 65% of the carbon-14 in a living snail.  How old is the site?  The half-life of carbon-14 is 5668 years.

Solution:  We know that 0.65 = 2^(-n/5668).
Hence log20.65 = -n/5668.  Solve this for n to get n = -(5668)log20.65 = 3522.6.  Therefore, the site is at least 3,522.6 years old.

EXERCISES

1.  A population of 1000 is expected to grow exponentially over the next ten years at 4% per year.  How large will the population be at the end of 10 years?

2.  A certain radioactive substance has a half-life of 30 seconds.  What percentage of an initial amount of this substance will remain after 10 seconds?

3.  If you invest \$5,000 in a money market fund at 15 percent interest compounded annually, how much will you have after 10 years?

4. Approximately how long will it take for money to accumulate to twice its value if it is invested at 8% compounded annually?

5. Suppose that you invest P dollars at r% compounded annually.  Write an expression for the amount accumulated after n years.

6. A certain radioactive element has a half-life of 400 years.  Starting with 50 milligrams there will be q(t) milligrams left after t years, where q(t) = 50(1/2)^kt..  How much will be left after 2257 years?

7. Bones from a human body were found to contain only 76% of the carbon-14 in living bones.  How long before did the person die?  The half-life of carbon-14 is 5668 years.

8. Suppose that you received a gift of \$10,000 this year and placed it in an fund that accrues interest at 10% compounded annually.  If you do not withdraw it until the year 2040, how much will you have then?

ANSWERS

1 = 1480.24

2 = 79%

3 \$20,227.80

4 9 years

5 P(1 + r/100)^n

6 1 milligram

7 2,244 years

8 \$452,593

Word problems/ applications of exponential and logarithmic functions