1. **State the problem:** Solve for $t$ in the equation $4614 = 3000 \left(1 + \frac{0.098}{2}\right)^{2t}$. 2. **Identify the formula:** This is a compound interest formula of the form $A = P \left(1 + \frac{r}{n}\right)^{nt}$ where: - $A$ is the amount after time $t$ - $P$ is the principal amount - $r$ is the annual interest rate - $n$ is the number of compounding periods per year - $t$ is the time in years 3. **Plug in known values:** $$4614 = 3000 \left(1 + \frac{0.098}{2}\right)^{2t}$$ 4. **Simplify inside the parentheses:** $$1 + \frac{0.098}{2} = 1 + 0.049 = 1.049$$ 5. **Rewrite the equation:** $$4614 = 3000 \times 1.049^{2t}$$ 6. **Divide both sides by 3000:** $$\frac{4614}{3000} = 1.049^{2t}$$ $$1.538 = 1.049^{2t}$$ 7. **Take the natural logarithm of both sides:** $$\ln(1.538) = \ln\left(1.049^{2t}\right)$$ 8. **Use logarithm power rule:** $$\ln(1.538) = 2t \ln(1.049)$$ 9. **Solve for $t$:** $$t = \frac{\ln(1.538)}{2 \ln(1.049)}$$ 10. **Calculate the values:** $$\ln(1.538) \approx 0.431$$ $$\ln(1.049) \approx 0.048$$ 11. **Final calculation:** $$t = \frac{0.431}{2 \times 0.048} = \frac{0.431}{0.096} \approx 4.49$$ **Answer:** The time $t$ is approximately $4.49$ years.