Question

A person invests 9000 dollars in a bank. The bank pays 6% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 14000 dollars?

Answer 1

`7.4\ years`

Explanation:

we know that

The compound interest formula is equal to

`A=P(1+(r)/(n))^(nt)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

`t=?\ years\\ P=\$9,000\\ r=0.06\\n=12\\ A=\$14,000`

substitute in the formula above

`14,000=9,000(1+(0.06)/(12))^(12t)`

`(14/9)=(1.005)^(12t)`

Apply log both sides

`log(14/9)=(12t)log(1.005)`

`t=log(14/9)/[(12)log(1.005)]`

`t=7.4\ years`