APPLICATIONS OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS

(Interest Rate Word Problems)

1. To solve an exponential or logarithmic word problems, convert the narrative to an equation and solve the equation.

Problem 4: If you invested $1,000 in an account paying an annual percentage rate (quoted rate) of 12%, compounded hourly (based on a bank year of 360 days), how much would you have in you account at the end of 1 year, 10 years, 20 years, 100 years?

Answer: 1 Year = $1,127.50, 10 years = $3,320.09, 20 years = $11,022.99, 100 years $162,741,385.44

Solution and Explanations:
Use the formula

displaymath45

where A is the balance at the end of a certain time period, P is the beginning investment of $1,000, and t is the number of years. The annual rate of 12% is converted to a hourly interest rate (based on a bank year of 360 days) since the compounding is hourly (24 x 360 times per year). Take the annual interest rate of 12% and divide by 8640 to obtain the hourly interest rate. The exponent is 8640t because there are 8,640 compounding periods in every year. Therefore, 8640t represents the number of compounding periods during t years.

To find the balance at the end of 1 year:
Substitute 1 for t in the equation tex2html_wrap_inline47 to derive A:

displaymath49

To find the balance at the end of 10 year:
Substitute 10 for t in the equation tex2html_wrap_inline47 to derive A:

displaymath53

To find the balance at the end of 20 year:
Substitute 20 for t in the equation tex2html_wrap_inline47 to derive A:

displaymath57

To find the balance at the end of 100 year:
Substitute 100 for t in the equation tex2html_wrap_inline47 to derive A:

displaymath61

If you would like to work another problem and check the answer and solution, click on Problem.

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Author: Nancy Marcus

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