# Precalculus : Application Problems

## Example Questions

### Example Question #1 : Application Problems

If you invest  into a savings account which earns an interest rate  per year, how much would it take for your deposit to double?

Explanation:

The equation of the value for this problem is

.

We can divide by R to get

.

We want to solve for n in this case, which is the amount of years. If we use the natural log on both sides and properties of logarithms, we get

.

If we solve for n, we get

### Example Question #1 : Application Problems

If you deposit  into a savings account which earns a  yearly interest rate, how much is in your account after two years?

Explanation:

Since we are investing for two years with a yearly rate of 5%, we will use the formula to calculate compound interest.

where

is the amount of money after time.

is the principal amount (initial amount).

is the interest rate.

is time.

Our amount after two years is:

### Example Question #1 : Application Problems

If you deposit  into a savings account which compounds interest every month, what is the expression for the amount of money in your account after  years if you earn a nominal interest rate of  compounded monthly?

Explanation:

Since  is the nominal interest rate compounded monthly we write the interest term as  as it is the effective monthly rate.

We compound for  years which is  months. Since our interest rate is compounded monthly our time needs to be in the same units thus, months will be the units of time.

Plugging this into the equation for compound interest gives us the expression:

### Example Question #4 : Application Problems

John opens a savings account and deposits  into it. This savings account gains  interest per year. After  years, John withdraws all the money, and deposits it into another savings account with  interest per year.  years later, John withdraws the money.

How much money does John have after this  year period? (Assume compound interest in both accounts)

Explanation:

Plugging our numbers into the formula for compound interest, we have:

.

So John has about  after the first three years.

After placing his money into the other savings account, he has

after  more years.

So John has accumulated about .

### Example Question #5 : Application Problems

Suppose you took out a loan  years ago that gains  interest. Suppose that you haven't made any payments on it yet, and right now you owe  on the loan. How much was the loan worth when you took it out?

Explanation:

The formula for the compund interest is as follows:

By substuting known values into the compound interest formula, we have:

.

From here, substitute known values.

Divide by

### Example Question #6 : Application Problems

How many years does it take for  to grow into  when the is deposited into a savings account that gains  annual compound interest?

years

years

years

years

Explanation:

To find the number of years required, we solve the compound interest formula for .

The formula is as follows:

Substitute known values.

Divide by .

Take the natural log of both sides.

Use the log power rule.

Divide by .

use a calculator to simplify.

### Example Question #7 : Application Problems

The amount of phosphorus present in a sample at a given time is given by the following equation:

Where  is in days,  is the amount of phosphorus after  days, and is the initial amount of phosphorus at the beginning of the first day. What percent of the initial amount of phosphorus is left after  days of decay?

Explanation:

The problem asks us for the percent that the amount of phosphorus after t days is of the original amount of phosphorus. If we think of this percentage with respect to the variables present in the equation, we can see that the following fraction expresses the amount of phosphorus after t days as a percentage of the initial amount of phosphorus:

So if this is the fraction we want to solve for, we should divide both sides of the equation by   to obtain this fraction on the left side of the equation:

We now have the fraction we want to solve for in terms of just one variable, , for which we plug in 25 days to find the percentage of phosphorus left of the initial amount after 25 days:

So after 25 days of decay, the amount of phosphorus is 47% of the initial amount.

### Example Question #1 : Application Problems

The exponential decay of an element is given by the following function:

Where  is the amount of the element left after  days, and is the initial amount of the element. If there are  kg of the element left after  days, what was the initial amount of the element?

kg

kg

kg

kg

kg

kg

Explanation:

The problem asks us for the initial amount of the element, so first let's solve our equation for :

The problem tells us that 25 days has passed, which gives us , and it also tells us the amount left after 25 days, which gives us . Now that we have our equation for , we can plug in the given values to find the initial amount of the element:

kg

### Example Question #2 : Application Problems

The exponential decay of an element is given by the function:

In this function,  is the amount of the element left after  days, and  is the initial amount of the element. If  of the element is left after seven days, how much of the element was there to begin with, rounded to the nearest kilogram?

Explanation:

To find the initial amount, you must rearrange the equation to solve for :

Divide both sides by :

Substituting in the values from the problem gives

### Example Question #1 : Application Problems

The exponential decay of an element is given by the function

In this function,  is the amount left after  days, and  is the initial amount of the element. What percent of the element is left after ten days, rounded to the nearest whole percent?