Mohal is going to invest \( \$ 830 \) and leave it in an account for 8 years. Assuming the interest is compounded daily, what interest rate, to the nearest hundredth of a percent, would be required in order for Mohal to end up with \( \$ 1,110 \) ?

Given: - Principal, \( P = 830 \) - Future Value, \( FV = 1110 \) - Time, \( t = 8 \) years - Compounding frequency, \( n = 365 \) (daily compounding) The future value formula for compounded interest is: \[ FV = P\left(1+\frac{r}{n}\right)^{nt} \] Substitute the given values: \[ 1110 = 830\left(1+\frac{r}{365}\right)^{365 \times 8} \] \[ 1110 = 830\left(1+\frac{r}{365}\right)^{2920} \] **Step 1. Divide both sides by 830:** \[ \frac{1110}{830} = \left(1+\frac{r}{365}\right)^{2920} \] \[ \frac{1110}{830} \approx 1.33735 \] Thus, \[ 1.33735 = \left(1+\frac{r}{365}\right)^{2920} \] **Step 2. Take the natural logarithm of both sides:** \[ \ln(1.33735) = \ln \left[\left(1+\frac{r}{365}\right)^{2920}\right] \] \[ \ln(1.33735) = 2920 \ln\left(1+\frac{r}{365}\right) \] **Step 3. Solve for \(\ln\left(1+\frac{r}{365}\right)\):** \[ \ln\left(1+\frac{r}{365}\right) = \frac{\ln(1.33735)}{2920} \] **Step 4. Exponentiate both sides to isolate \(1+\frac{r}{365}\):** \[ 1+\frac{r}{365} = \exp\left(\frac{\ln(1.33735)}{2920}\right) \] **Step 5. Solve for \( r \):** \[ \frac{r}{365} = \exp\left(\frac{\ln(1.33735)}{2920}\right) - 1 \] \[ r = 365\left[\exp\left(\frac{\ln(1.33735)}{2920}\right) - 1\right] \] **Step 6. Compute the numerical value:** First, compute \(\ln(1.33735)\): \[ \ln(1.33735) \approx 0.2904 \] Then, \[ \frac{0.2904}{2920} \approx 0.00009945 \] Now, calculate the exponential: \[ \exp(0.00009945) \approx 1.00009945 \quad \text{(since for small } x, \exp(x) \approx 1+x \text{)} \] Thus, \[ \frac{r}{365} \approx 1.00009945 - 1 = 0.00009945 \] \[ r \approx 365 \times 0.00009945 \approx 0.036304 \quad \text{(in decimal form)} \] To convert to a percentage: \[ r \approx 0.036304 \times 100 \approx 3.63\% \] The required interest rate, to the nearest hundredth of a percent, is \(3.63\%\).