Question

Find the time required for an investment of $3000 to grow to $9000 at an interest rate of 6.5% per year, compounded quarterly. (Round your answer to two decimal places.)

Answer 1

Answer:

$t=17.04\ years$

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  

$P=\ 3,000\\A=\ 9,000\\ r=6.5\%=6.5/100=0.065\\n=4$  

substitute in the formula above

$9,000=3,000(1+\frac{0.065}{4})^{4t}$  

solve for t

Simplify

$3=(1.01625)^{4t}$  

Apply log both sides

$log(3)=log[(1.01625)^{4t}]$  

Apply property of logarithms

$log(3)=(4t)log(1.01625)$  

$t=log(3)/[4log(1.01625)]$  

$t=17.04\ years$