Question

Bentley invested $430 in an account paying an interest rate of 2.5% compounded continuously. Assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach $510?

Answer 1

`~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&\$510\\ P=\textit{original amount deposited}\dotfill & \$430\\ r=rate\to 2.5\%\to (2.5)/(100)\dotfill &0.025\\ t=years \end{cases}`

`510=430e^(0.025\cdot t)\implies \cfrac{510}{430}=e^(0.025t)\implies \cfrac{51}{43}=e^(0.025t) \\\\\\ \log_e\left( \cfrac{51}{43} \right)=\log_e\left( e^(0.025t) \right)\implies \ln\left( \cfrac{51}{43} \right)=0.025t \\\\\\ \cfrac{\ln\left( (51)/(43) \right)}{0.025}=t\implies 6.83\approx t\implies \stackrel{\textit{rounded up}}{7=t}`