Gianna is going to invest \( \$ 700 \) and leave it in an account for 14 years. Assuming the interest is compounded daily, what interest rate, to the nearest tenth of a percent, would be required in order for Gianna to end up with \( \$ 1,010 \) ?

**Step 1. Write the compound interest formula** We use the formula for compound interest: \[ A = P\left(1 + \frac{r}{n}\right)^{nt} \] where \(P = 700\) (initial investment), \(A = 1010\) (final amount), \(t = 14\) years, and \(n = 365\) (number of compounding periods per year). --- **Step 2. Substitute the known values** Substitute the given numbers into the formula: \[ 1010 = 700\left(1 + \frac{r}{365}\right)^{365 \times 14} \] Simplify the exponent: \[ 365 \times 14 = 5110 \] Thus, the equation becomes: \[ 1010 = 700\left(1 + \frac{r}{365}\right)^{5110} \] --- **Step 3. Isolate the compound factor** Divide both sides by 700: \[ \frac{1010}{700} = \left(1 + \frac{r}{365}\right)^{5110} \] Simplify the fraction: \[ \frac{1010}{700} \approx 1.442857 \] So we have: \[ 1.442857 = \left(1 + \frac{r}{365}\right)^{5110} \] --- **Step 4. Take the natural logarithm of both sides** Apply the natural logarithm: \[ \ln(1.442857) = \ln\left[\left(1 + \frac{r}{365}\right)^{5110}\right] \] Use the logarithmic power rule: \[ \ln(1.442857) = 5110 \cdot \ln\left(1 + \frac{r}{365}\right) \] Solve for \(\ln\left(1 + \frac{r}{365}\right)\): \[ \ln\left(1 + \frac{r}{365}\right) = \frac{\ln(1.442857)}{5110} \] --- **Step 5. Solve for \(\frac{r}{365}\)** Exponentiate both sides to remove the logarithm: \[ 1 + \frac{r}{365} = \exp\left(\frac{\ln(1.442857)}{5110}\right) \] Subtract 1 from both sides: \[ \frac{r}{365} = \exp\left(\frac{\ln(1.442857)}{5110}\right) - 1 \] --- **Step 6. Solve for \(r\)** Multiply both sides by 365: \[ r = 365 \left[\exp\left(\frac{\ln(1.442857)}{5110}\right) - 1\right] \] --- **Step 7. Compute the values** First, calculate \(\ln(1.442857)\): \[ \ln(1.442857) \approx 0.3666 \] Then compute: \[ \frac{0.3666}{5110} \approx 7.18 \times 10^{-5} \] Since \(7.18 \times 10^{-5}\) is very small, we can use the approximation \(\exp(x) \approx 1 + x\) for small \(x\): \[ \exp\left(7.18 \times 10^{-5}\right) \approx 1 + 7.18 \times 10^{-5} \] Thus, \[ r \approx 365 \times 7.18 \times 10^{-5} \approx 0.0262 \] --- **Step 8. Express the interest rate as a percentage** Converting \( r \) to a percentage: \[ r \approx 0.0262 \times 100\% \approx 2.62\% \] Rounded to the nearest tenth of a percent: \[ r \approx 2.6\% \] --- **Answer:** The required interest rate is approximately \(2.6\%\).