Question

Deshaun needs 3000 for a future project. He can invest 2000 now at an annual rate of 9.5%, compounded semiannually. Assuming that no withdrawals are made, how long will it take for him to have enough money for his project? Do not round any intermediate computations, and round your answer to the nearest hundredth.

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\$3000\\ P=\textit{original amount deposited}\dotfill &\$2000\\ r=rate\to 9.5\%\to (9.5)/(100)\dotfill &0.095\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannually, thus twice} \end{array}\dotfill &2\\ t=years \end{cases}`

`3000=2000\left(1+(0.095)/(2)\right)^(2\cdot t) \implies \cfrac{3000}{2000}=1.0475^(2t)\implies \cfrac{3}{2}=1.0475^(2t) \\\\\\ \log\left( \cfrac{3}{2} \right)=\log(1.0475^(2t))\implies \log\left( \cfrac{3}{2} \right)=t\log(1.0475^(2)) \\\\\\ \cfrac{\log\left( (3)/(2) \right)}{\log(1.0475^(2))}=t\implies 4.37\approx t\qquad \textit{about 4 years and 4 months and a half}`