Question

A person invests 7500 dollars in a bank. The bank pays 6% interest compounded semi-annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 16200 dollars?

Answer 1

Answer:

$13.0\ years$  

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  

$t=?\ years\\ P=\ 7,500\\ r=0.06\\n=2\\ A=\ 16,200$  

substitute in the formula above  

$16,200=7,500(1+\frac{0.06}{2})^{2t}$  

$(2.16)=(1.03)^{2t}$  

Apply log both sides

$log(2.16)=(2t)log(1.03)$  

$t=log(2.16)/[(2)log(1.03)]$  

$t=13.0\ years$