Question

a person invests 9000 dollars in the back. The back pats 4.75% interest compounded quarterly. To the tenth of a year, how long must a person leave the money in the back until it reaches 11800 dollars

Answer 1

Answer:

5.7 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  

$t=?\ years\\ P=\ 9,000\\A=\ 11,800\\r=4.75\%=4.75/100=0.0475\\n=4$

substitute in the formula above

$11,800=9,000(1+\frac{0.0475}{4})^{4t}$  

Solve for t

$\frac{11.8}{9} =(1.011875)^{4t}$  

Applying property of exponents

$\frac{11.8}{9} =(1.011875^{4})^{t}$  

Applying log both sides

$log(\frac{11.8}{9})=log[(1.011875^{4})^{t}]$  

$log(\frac{11.8}{9})=(t)log[(1.011875^{4})]$  

$t=log(\frac{11.8}{9})/log[(1.011875^{4})]$  

$t=5.7\ years$