Question

A person invests 9000 dollars in a bank. The bank pays 6% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 14000 dollars?

Answer 1

Answer:

$7.4\ years$  

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  

$t=?\ years\\ P=\ 9,000\\ r=0.06\\n=12\\ A=\ 14,000$  

substitute in the formula above  

$14,000=9,000(1+\frac{0.06}{12})^{12t}$  

$(14/9)=(1.005)^{12t}$  

Apply log both sides

$log(14/9)=(12t)log(1.005)$  

$t=log(14/9)/[(12)log(1.005)]$  

$t=7.4\ years$  

Answer 2

Answer:

19.1 years

Step-by-step explanation:

i did the test