Question

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

Answer 1

Answer:

$r=3.93\%$

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

$A=P(1+\frac{r}{n})^{nt}$  

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

$2x=x(1+\frac{r}{1})^{1*18}$  

$2=(1+r})^{18}$  

Elevated both sides to 1/18

$2^{\frac{1}{18}} =1+r$

$r=2^{\frac{1}{18}} -1$

$r=0.0393$

convert to percentage

Multiply by 100

$r=3.93\%$