Question

If $5200 is invested at a rate of 8.6% compounded quarterly, how long will it take, to the nearest hundredth of a year, until the investment is worth $11,600

Answer 1

Answer:

$t= 9.43\ years$

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  

$t=?\ years\\ P=\ 5,200\\ r=8.6\%=8.6\100=0.086\\n=4\\A=\ 11,600$  

substitute in the formula above

$11,600=5,200(1+\frac{0.086}{4})^{4t}$  

$11,600=5,200(1.0215)^{4t}$  

$(11,600/5,200)=(1.0215)^{4t}$  

Applying log both sides

$log(11,600/5,200)=log(1.0215)^{4t}$  

applying property of logarithms

$log(11,600/5,200)=(4t)log(1.0215)$  

$t=log(11,600/5,200)/[(4)log(1.0215)]$

$t= 9.43\ years$