Question

A person invests 1000 dollars in a bank. The bank pays 6% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 2100 dollars?

Answer 1

For this case we have the following information given:

A= 1000

r= 0.06

n = 1

A= 2100

So we can set up the following formula:

`2100=1000(1+(0.06)/(1))^t`

We can solve for t on this way:

`\ln ((2100)/(1000))=t\cdot\ln (1.06)`

Solving for t we got:

`t=(\ln (2.1))/(\ln (1.06))=12.73`

Rounded to the nesrest tenth we got:

12.7 years