Question

How long will it take money to double if it is invested at the following rates?

(A) 6% compounded monthly
(B) 10.2% compounded monthly
(A) years
(Round to two decimal places as needed.)
(B) years
(Round to two decimal places as needed.)

Answer 1

11.58 years

6.82 years

Explanation:

Compound interest can be represented as:
`A=P(1+(r)/(n))^(nt)` where P = initial amount, r=interest rate, n=compounds in each time unit, t = time (usually years)

If you think about it the:
`(1+(r)/(n)})^(nt)` represents the overall interest being applied to the principal amount.

This would have to be equal to 2, for it to be doubled, so let's set that equal to 2

`2=(1+(r)/(n))^(nt)`

Now let's solve for "t", first let's take the log of both sides

`log(2)=log([1+(r)/(n)]^(nt))`

Now let's use log properties, specifically the exponent one to rewrite

`log(2)=nt *log(1+(r)/(n))`

Now divide both sides by the log(1 + r/n) and n

`t=(log(2))/(n*log(1+(r)/(n)))`

Our "n" is going to be the same for both equations, 12 since it's compounded monthly

The interest will be a bit different, but in the first case it will be 6% which is 0.06

`t=(log(2))/(12*log(1+(0.06)/(12)))`

simplifying this we get

`t\approx11.581`

Rounding this we get:
`t=11.58`

Now let's do this to the other equation:

`t=(log(2))/(12*log(1+(0.102)/(12)))`

simplifying this we get:

`t\approx 6.824`

approximately 6.82