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Suppose that $6493 is invested at an interest rate of 42% per year, compounded continuously. What is the account balance after 5 years?


Using \(\displaystyle P=P_0e^{kt}\) gives \(\displaystyle P=6493e^{\frac{21 t}{50}}\). Evaluating at \(\displaystyle t=5\) gives \(\displaystyle P=53023\) dollars.

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\begin{question}Suppose that \$6493 is invested at an interest rate of 42\% per year, compounded continuously. What is the account balance after 5 years? 
    \soln{4.5cm}{Using $P=P_0e^{kt}$ gives $P=6493e^{\frac{21 t}{50}}$. Evaluating at $t=5$ gives $P=53023$ dollars.}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas
<p> <p>Suppose that &#36;6493 is invested at an interest rate of 42&#37; per year, compounded continuously. What is the account balance after 5 years? </p> </p>
HTML for Canvas
<p> <p>Using  <img class="equation_image" title=" \displaystyle P=P_0e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20P%3DP_0e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle P=P_0e^{kt} " data-equation-content=" \displaystyle P=P_0e^{kt} " />  gives  <img class="equation_image" title=" \displaystyle P=6493e^{\frac{21 t}{50}} " src="/equation_images/%20%5Cdisplaystyle%20P%3D6493e%5E%7B%5Cfrac%7B21%20t%7D%7B50%7D%7D%20" alt="LaTeX:  \displaystyle P=6493e^{\frac{21 t}{50}} " data-equation-content=" \displaystyle P=6493e^{\frac{21 t}{50}} " /> . Evaluating at  <img class="equation_image" title=" \displaystyle t=5 " src="/equation_images/%20%5Cdisplaystyle%20t%3D5%20" alt="LaTeX:  \displaystyle t=5 " data-equation-content=" \displaystyle t=5 " />  gives  <img class="equation_image" title=" \displaystyle P=53023 " src="/equation_images/%20%5Cdisplaystyle%20P%3D53023%20" alt="LaTeX:  \displaystyle P=53023 " data-equation-content=" \displaystyle P=53023 " />  dollars.</p> </p>