1. **State the problem:** We are given that $\sin(\ln x) = \frac{1}{4}$ and $\sin(\ln y) = \frac{1}{7}$ with $0 < \ln x < \frac{\pi}{2}$ and $0 < \ln y < \frac{\pi}{2}$. We need to find $\sin(\ln(xy))$ using the sum-of-angle sine formula and round the answer to three decimal places. 2. **Recall the sum-of-angle sine formula:** $$\sin(a+b) = \sin a \cos b + \cos a \sin b$$ Here, let $a = \ln x$ and $b = \ln y$, so $$\sin(\ln(xy)) = \sin(\ln x + \ln y) = \sin(\ln x) \cos(\ln y) + \cos(\ln x) \sin(\ln y)$$ 3. **Find $\cos(\ln x)$ and $\cos(\ln y)$ using the Pythagorean identity:** Since $\sin^2 \theta + \cos^2 \theta = 1$, we have $$\cos(\ln x) = \sqrt{1 - \sin^2(\ln x)} = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$$ Similarly, $$\cos(\ln y) = \sqrt{1 - \sin^2(\ln y)} = \sqrt{1 - \left(\frac{1}{7}\right)^2} = \sqrt{1 - \frac{1}{49}} = \sqrt{\frac{48}{49}} = \frac{\sqrt{48}}{7} = \frac{4\sqrt{3}}{7}$$ 4. **Substitute all values into the sum formula:** $$\sin(\ln(xy)) = \left(\frac{1}{4}\right) \left(\frac{4\sqrt{3}}{7}\right) + \left(\frac{\sqrt{15}}{4}\right) \left(\frac{1}{7}\right) = \frac{\sqrt{3}}{7} + \frac{\sqrt{15}}{28}$$ 5. **Calculate the numerical value:** $$\frac{\sqrt{3}}{7} \approx \frac{1.732}{7} \approx 0.2474$$ $$\frac{\sqrt{15}}{28} \approx \frac{3.873}{28} \approx 0.1383$$ Adding these, $$0.2474 + 0.1383 = 0.3857$$ 6. **Round to three decimal places:** $$\sin(\ln(xy)) \approx 0.386$$