Question

A person invest 9500 dollars in a bank. The bank pays 4.5% interest compounded daily. To the nearest tenth of a year , how long must the person leave the money in the bank until it reaches 19300 dollars ?

Answer 1

15.8 years

Explanation:

Answer 2

`t=15.6\ 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=\$9,500\\A=\$19,300\\r=4.5\%=4.5/100=0.045\\n=365`

substitute in the formula above

`19,300=9,500(1+(0.045)/(365))^(365t)`

solve for t

simplify

`(19,300/9,500)=((365.045)/(365))^(365t)`

Apply log both sides

`log(19,300/9,500)=log[((365.045)/(365))^(365t)]`

Apply property of logarithms in the right side

`log(19,300/9,500)=(365t)log((365.045)/(365))`

`t=log(19,300/9,500)/[(365)log((365.045)/(365))]`

`t=15.6\ years`