rou deposit \( \$ 3000 \) into an account that earns \( 5 \% \) Thempounded annually. A friend deposits \( \$ 2250 \) into an Thecount that earns \( 4.95 \% \) annual interest, compounded continuously. Will your friend's balance ever equal ours? If so, when? If not, enter NEVER. nearest whole year). help (numbers)

We set up the two growth formulas. For your account (compounded annually), the balance after \( t \) years is \[ A(t) = 3000(1.05)^t. \] For your friend’s account (compounded continuously), the balance after \( t \) years is \[ B(t) = 2250\,e^{0.0495\,t}. \] We want to find \( t \) such that \[ 3000(1.05)^t = 2250\,e^{0.0495\,t}. \] 1. Divide both sides by \( 2250 \): \[ \frac{3000}{2250}(1.05)^t = e^{0.0495\,t}. \] Simplify \( \frac{3000}{2250} \) to obtain: \[ \frac{4}{3}(1.05)^t = e^{0.0495\,t}. \] 2. Take the natural logarithm of both sides: \[ \ln\left(\frac{4}{3}(1.05)^t\right) = \ln\left(e^{0.0495\,t}\right). \] Use the logarithm properties: \[ \ln\left(\frac{4}{3}\right) + t\,\ln(1.05) = 0.0495\,t. \] 3. Isolate the \( t \)-terms: \[ \ln\left(\frac{4}{3}\right) = t\left(0.0495 - \ln(1.05)\right). \] 4. Solve for \( t \): \[ t = \frac{\ln\left(\frac{4}{3}\right)}{0.0495 - \ln(1.05)}. \] 5. Compute the numerical values: - \(\ln\left(\frac{4}{3}\right) \approx 0.287682072\), - \(\ln(1.05) \approx 0.048790164\). Then, \[ 0.0495 - 0.048790164 \approx 0.000709836, \] and \[ t \approx \frac{0.287682072}{0.000709836} \approx 405.27. \] Rounding to the nearest whole year, the friend’s balance will equal your balance in approximately \[ \boxed{405} \text{ years}. \] Thus, yes, your friend’s balance will eventually equal yours after about 405 years.