Amira is going to invest $580 and leave it in an account for 17 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Amira to end up with $840?

To solve this problem, we will use the formula for continuous compounding interest, which is:

$A = P e^{rt}$

where:

• $A$ is the amount of money accumulated after n years, including interest.
• $P$ is the principal amount (the initial amount of money).
• $r$ is the annual interest rate (in decimal form).
• $t$ is the time the money is invested for, in years.
• $e$ is the base of the natural logarithm, approximately equal to 2.71828.

Given:

• $A = 840$
• $P = 580$
• $t = 17$

We need to find $r$.

1. Substitute the given values into the formula:

$840 = 580 e^{17r}$

2. Divide both sides by 580 to isolate the exponential term:

$\frac{840}{580} = e^{17r}$

3. Simplify the fraction:

$\frac{840}{580} \approx 1.4483$

So,

$1.4483 = e^{17r}$

4. Take the natural logarithm (ln) of both sides to solve for $r$:

$\ln(1.4483) = \ln(e^{17r})$

5. Use the property of logarithms that $\ln(e^x) = x$:

$\ln(1.4483) = 17r$

6. Calculate $\ln(1.4483)$:

$\ln(1.4483) \approx 0.3701$

So,

$0.3701 = 17r$

7. Solve for $r$ by dividing both sides by 17:

$r = \frac{0.3701}{17}$

$r \approx 0.0218$

8. Convert the decimal to a percentage by multiplying by 100:

$r \approx 2.18\%$

Therefore, the interest rate required for Amira to end up with $840 after 17 years, compounded continuously, is approximately 2.18%.