1. **Problem:** Find the year when the population of West Goma equals the population of East Goma given: $$f(x) = 16.4e^{0.0002x}$$ $$g(x) = 13.7e^{0.0134x}$$ where $x=0$ corresponds to September 1998. 2. **Set the populations equal:** $$16.4e^{0.0002x} = 13.7e^{0.0134x}$$ 3. **Divide both sides by $13.7e^{0.0002x}$ to isolate the exponential term:** $$\frac{16.4e^{0.0002x}}{13.7e^{0.0002x}} = \frac{13.7e^{0.0134x}}{13.7e^{0.0002x}}$$ $$\frac{\cancel{16.4}e^{\cancel{0.0002x}}}{\cancel{13.7}e^{\cancel{0.0002x}}} = e^{0.0134x - 0.0002x}$$ Simplifies to: $$\frac{16.4}{13.7} = e^{0.0132x}$$ 4. **Take the natural logarithm of both sides:** $$\ln\left(\frac{16.4}{13.7}\right) = \ln\left(e^{0.0132x}\right)$$ $$\ln\left(\frac{16.4}{13.7}\right) = 0.0132x$$ 5. **Solve for $x$:** $$x = \frac{\ln\left(\frac{16.4}{13.7}\right)}{0.0132}$$ Calculate numerator: $$\ln\left(\frac{16.4}{13.7}\right) = \ln(1.19635) \approx 0.179$$ Then: $$x = \frac{0.179}{0.0132} \approx 13.56$$ 6. **Interpretation:** Since $x=0$ is September 1998, the populations are equal approximately $13.56$ years later. Add $13.56$ years to 1998: $$1998 + 13.56 \approx 2011.56$$ So, around mid-2011. **Final answer:** The populations of West Goma and East Goma will be equal around the year **2011**.