Question

What rate of interest compounded annually is required to double an investment
in 18 ​years?

Answer 1

`r=3.93\%`

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=18\ years\\ P=x\\A=2x\\ r=?\\n=1`

substitute in the formula above

`2x=x(1+(r)/(1))^(1*18)`

`2=(1+r})^(18)`

Elevated both sides to 1/18

`2^{(1)/(18)} =1+r`

`r=2^{(1)/(18)} -1`

`r=0.0393`

convert to percentage

Multiply by 100

`r=3.93\%`