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

5.7 years

Explanation:

we know that

The compound interest formula is equal to

`A=P(1+(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+(0.0475)/(4))^(4t)`

Solve for t

`(11.8)/(9) =(1.011875)^(4t)`

Applying property of exponents

`(11.8)/(9) =(1.011875^(4))^(t)`

Applying log both sides

`log((11.8)/(9))=log[(1.011875^(4))^(t)]`

`log((11.8)/(9))=(t)log[(1.011875^(4))]`

`t=log((11.8)/(9))/log[(1.011875^(4))]`

`t=5.7\ years`