Question

$9,800.00 is invested at an annual rate of 11.6%, compounded continuously. How long will it take for that investment to grow to a value of 149,000.00? Round your answer (in years) to two decimal places at the END of your calculations.

Answer 1

`~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$149000\\ P=\textit{original amount deposited}\dotfill & \$9800\\ r=rate\to 11.6\%\to (11.6)/(100)\dotfill &0.116\\ t=years \end{cases} \\\\\\ 149000=9800e^(0.116\cdot t)\implies \cfrac{149000}{9800}=e^(0.116\cdot t)\implies \cfrac{745}{49}=e^(0.116t)`

`\log_e\left( \cfrac{745}{49} \right)=\log_e\left( e^(0.116t) \right)\implies \log_e\left( \cfrac{745}{49} \right)=0.116t \\\\\\ \cfrac{\ln\left( (745)/(49) \right)}{0.116}=t\implies 23.46\approx t`

Answer 2

Explanation:

First, convert R as a percent to r as a decimal

r = R/100

r = 3.875/100

r = 0.03875 rate per year,

Then solve the equation for A

A = P(1 + r/n)nt

A = 10,000.00(1 + 0.03875/12)(12)(7.5)

A = 10,000.00(1 + 0.003229167)(90)

A = $13,366.37

Summary:

The total amount accrued, principal plus interest, with compound interest on a principal of $10,000.00 at a rate of 3.875% per year compounded 12 times per year over 7.5 years is $13,366.37.