3. tan invests $465 in an account that earns 3.1% annual interest compounded continuously. find when the…

Explanation:

Step1: Recall continuous - compounding formula

The formula for continuous - compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $t$ is the time in years. Given $P = 465$, $r=0.031$, and $A = 2400$. Substitute these values into the formula: $2400=465e^{0.031t}$.

Step2: Isolate the exponential term

Divide both sides of the equation by 465: $\frac{2400}{465}=e^{0.031t}$. Simplify $\frac{2400}{465}=\frac{160}{31}$, so $\frac{160}{31}=e^{0.031t}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{160}{31})=\ln(e^{0.031t})$. Since $\ln(e^{x}) = x$, the right - hand side simplifies to $0.031t$. So, $\ln(\frac{160}{31}) = 0.031t$.

Step4: Solve for $t$

We know that $\ln(\frac{160}{31})\approx\ln(160)-\ln(31)\approx5.0752 - 3.4339=1.6413$. Then $t=\frac{\ln(\frac{160}{31})}{0.031}=\frac{1.6413}{0.031}\approx52.945$.

Answer:

Approximately $52.95$ years.