In another MathFAQ, I looked at how you can find the rate in the compound interest formula. Now let’s look at an example where we solve for the number of years n. This problem is different because what we are looking for appears in a power.

Problem Suppose $5000 is deposited in an account that earns 2% compound interest that is done annually. In how many years will there be $6000 in the account.

Solution This problem requires the use of the compound interest formula,

This formula applies when interest is earned on an annual basis and the interest is earned once a year.

Let’s look at the quantities in the problem statement:

$5000 is deposited in an account > P = 5000

that earns 2% compound interest that is done annually > r = 0.02

Will there be $6000 in the account > A = 6000

Putting these values into the formula above gives us

Unlike other problems where we solve for P or r, here we need to solve for the power in the right hand side, n. Solving for a value in the power requires the property of logarithms, log(y^{x}) = x logy. It allows us to move the n in the power and change it to a multiplier. But before we can apply this property, we isolate the factor containing the n:

Now take the logarithm of both sides of the equation:

This gives us

or n ≈ 9.21 years.

In WolframAlpha, we could evaluate the logs as follows.

In a previous FAQ, I looked at an equation for modeling the growth of the population on Gilligan’s Island,

A = 7 (1+0.02)^{n}

This equation assumed that the initial was 7 in 1964 and grew by 2% in each year after that. That means that if you put in n = 14, you will have multiplied the initial population by 1.02 a total of 14 times (1978).

How many years would you need to go to double the original population?

Those of you who are fans of classic TV will recognize this as introduction to the sitcom “Gilligan’s Island”. This series ran from September 26, 1964 to September 4, 1967. The series revolved around 7 castaways (4 men and 3 women) marooned on an island somewhere near Hawaii.

Over the course of three seasons, the series followed attempts to leave the island, visitors to the island and general incompetence in getting rescued. When the series was cancelled in 1967, the castaways were never rescued.

In 1978, a made for TV movie called “Rescue from Gilligan’s Island” aired in which the castaways were rescued. However, at the end of this movie they decided to go on a reunion cruise and became stranded on the same island again after another freak storm. In 1979, another made for TV movie called “The Castaways on Gilligan’s Island” aired in which they were rescued once again. This time they decide to convert the island to a resort. It was hoped that this premise would generate a new series, but this never happened.

In 1981, a second sequel was created, “The Harlem Globetrotters on Gilligan’s Island” in which nefarious forces plotted to take over the island and the castaways are saved by the Harlem Globetrotter.

Although this series began before I was born, I watched years and years of reruns throughout my childhood. I must have watched each of the 98 episodes several times each. But there was one question that nagged at me. If there were two women on the island of child bearing age (Ginger and Mary Ann), why didn’t the population grow? Why didn’t nature take its course and lead to new castaways?

Let’s find out how many castaways there should have been in 1978, 14 years after they were originally shipwrecked.

To answer this question, let’s assume that the population on the island is an example of exponential growth. This may or may not be a good choice due to the small size of the population. However, with this assumption the amount of peoplec A on the island n years after 1964 will be given by the equation

A = P (1+r)^{n}

In this situation, P is the initial amount of people on the island and r is the annual growth rate in percent per year.

Since the population of Gilligan’s Island was initially 7, we’ll set P = 7. For the growth rate r, we’ll use a fairly conservative rate of 2% per year. This is about what the world birth rate is. Around the world birth rates vary from a little less than 1% (China) to around 5% (African countries). With these values, we model the population by

A = 7 (1+0.02)^{n}

To predict the population in 1978, I’ll find the value at n = 14.

A = 7 (1+0.02)^{14} = 7(1.02)^{14 }=^{ }9.24

According to this growth model, there should have been about 9 people on the island in 1978.

This equation assumed that the initial was 7 in 1964 and grew by 2% in each year after that. That means that if you put in n = 14, you will have multiplied the initial population by 1.02 a total of 14 times (1978).

How many years would you need to go to double the original population?

Since the original population was 7, double that number is 14. So if I put that number into A, I need to solve

14 = 7 (1.02)^{n}

for the number of years n after 1964.

Start by dividing both sides by 7 to isolate the piece with the exponent in it,

2 = (1.02)^{n}

Now take the logarithm of both sides,

log(2) = log((1.02)^{n})

The reason we need to use logarithms is that powers in front of logarithms may be moved to the front,

log(2) = n log(1.02)

Now divide both sides by log(1.02),

^{log(2)}/_{log(1.02)} = n

Using a calculator to compute the logs gives a value of n of about 35. So in 1964 + 35 or 1995, the population would have increased from 7 to 14 according to the equation.

Suppose 5000 dollars is deposited in an account that earns 2% compound interest that is done annually. In how many years will there be 6000 dollars in the account.

This problem requires the use of the compound interest formula,

This formula applies when interest is earned on an annual basis and the interest is earned once a year.

Let’s look at the quantities in the problem statement:

5000 dollars is deposited in an account > P = 5000

that earns 2% compound interest that is done annually > r = 0.02

Will there be 6000 dollars in the account > A = 6000

Putting these values into the formula above gives us

Unlike other problems where we solve for P or r, here we need to solve for the power in the right hand side, n. Solving for a value in the power requires the property of logarithms, log(y^{x}) = x logy. It allows us to move the n in the power and change it to a multiplier. But before we can apply this property, we isolate the factor containing the n:

Now take the logarithm of both sides of the equation:

This gives us

or n ≈ 9.21 years.

In WolframAlpha, we could evaluate the logs as follows.

After a product is released or an advertising campaign is finished, sales usually drop off. Often this decrease is modeled with an exponential function. This example shows how to find when the sales have dropped a a predetermined level. This requires us to convert from exponential to logarithm…a key skill from Section 5.2.

Problem Monthly sales of a Blue Ray player are approximately

where t is the number of months the Blue Ray player has been on the market.

a. Find the initial sales.

Solution The initial sales occur at t = 0. The corresponding sales are

or 250,000 units.

b. In how many months will sales reach 500,000 units?

Solution Set S(t) equal to 500 and solve for t.

c. Will sales ever reach 1000 thousand units?

Solution Follow steps similar to part b.

Since the logarithm of zero is not defined, sales will never be 1000 thousand units.

d. Is there a limit for sales?

Solution To help us answer this question, let’s look at a graph of S(t).

Examining the graph, it appears that the sales are getting closer and closer to 1000 units, but never quite get there (part c). So the limit for sales is 1000 thousand units or 1,000,000 units. This is due to the fact that as t increases, e^{–t} gets smaller and smaller so all that is left from S(t) is 1000.