Question

a person invest 9500 dollars in the bank. the bank pays 4.5% interest compounded quarterly. to the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 22100 dollars?

Answer 1

Step 1

State the compound interest formula

`A=P(1+(r)/(n))^(nt)`

where;

`\begin{gathered} P=\text{ \$}9500 \\ r=(4.5)/(100)=0.045 \\ n=4 \\ A=\text{ \$22100} \end{gathered}`

Step 2

Find how long must the person leave the money in the bank until it reaches 22100 dollars

`\begin{gathered} 22100=9500(1+(0.045)/(4))^(4* t) \\ 22100=9500((809)/(800))^(4t) \\ (9500\left((809)/(800)\right)^(4t))/(9500)=(22100)/(9500) \\ \left((809)/(800)\right)^(4t)=(221)/(95) \\ 4t\ln \left((809)/(800)\right)=\ln \left((221)/(95)\right) \\ t=(\ln \left((221)/(95)\right))/(4\ln \left((809)/(800)\right)) \\ t=18.86724 \\ t\approx18.9\text{ years to the nearest tenth of a year} \end{gathered}`

Answer;

`t=18.9\text{ years to the nearest tenth of a year}`