Exponential Equations: Compound Interest Application

One of the most common applications of the exponential functions is the calculation of compound and continuously compounded interest. This discussion will focus on the compound interest application.

The formula for compound interest is:

COMPOUND INTEREST FORMULA

$A = P {(1 + \frac{r}{n})}^{n t}$

Where A is the account balance, P the principal or starting value, r the annual interest rate as a decimal, n the number of compoundings per year and t the time in years.

Let's solve a few compound interest problems.

Antonin opened a savings account with $700. If the annual interest rate is 7.5%, what will the account balance be after 10 years?


Step 1: Identify the known variables. Remember that the rate must be in decimal form and n is the number of compoundings per year. Since this situation has an annual interest rate there is only 1 compounding per year.	A = ? Account balance P = $700 Starting value r = 0.075 Decimal form n = 1 No. compound. t = 10 No. of years
Step 2: Substitute the known values.	$A = 700 {(1 + \frac{0.075}{1})}^{(1) (10)}$
Step 3: Solve for A.	$A = 700 {(1 + \frac{0.075}{1})}^{(1) (10)}$ Original A = 700(1.075)¹⁰ Simplify A = $1442.72 Multiply

Example 1: After 5 years of interested payments of

5 \frac{1}{2}

% compounded quarterly, an account has $5046.02. What was the principal?

Step 1: Identify the known variables.

Remember that the rate must be in decimal form and n is the number of compoundings per year.

Since this situation has quarterly compounding there are 4 compoundings per year.

A = $5046.02 Account balance

P = ? Principal

r = 0.055 Decimal form

n = 4 No. compound.

t = 5 No. of years

Step 2: Substitute the known values.

$5046.02 = P {(1 + \frac{0.055}{4})}^{(4) (5)}$

Step 3: Solve for P.

5046.02 = P(1.01375)²⁰ Original

$\frac{5046.02}{{1.01375}^{20}} = P$ Divide

P = $3840.00

Example 2: A college fund is started for Ashton on his fifth birthday. The initial investment of $2500 is compounded bimonthly at a rate of 9%. How old will Ashton be when the account balanced has quadrupled?

Step 1: Identify the known variables. Remember that the rate must be in decimal form and n is the number of compoundings per year. Since this situation has bimonthly, twice a month, compounding there are 24 compoundings per year.	A = 4 x $2500 Account balance P = $2500 Principal r = 0.09 Decimal form n = 24 No. compound. t = ? No. of years
Step 2: Substitute the known values.	$10, 000 = 2500 {(1 + \frac{0.09}{24})}^{(24) (t)}$
Step 3: Solve for t.	10,000 = 2500(1.00375)^24t Original 4 = (1.00375)^24t Divide log_1.00375 4 = log_1.00375 (1.00375)^24t Log log_1.00375 4 = 24t Inverse $\frac{l o g_{1.00375} 4}{24} = t$ Divide $\frac{1}{24} \times \frac{l o g 4}{\log 1.00375} = t$ Change base $t \approx 15.4$
Step 4: Solve for Ashton's age.	$5 + 15.4 = 20.4 \approx 20$ years old

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