1. **Problem:** Solve the simultaneous equations: $$\log_2 x + \log_2 y = 3$$ $$\log_7 x = 2$$ 2. **Step 1: Use the properties of logarithms.** Recall that $\log_a b + \log_a c = \log_a (bc)$. So, the first equation becomes: $$\log_2 (xy) = 3$$ 3. **Step 2: Convert logarithmic equations to exponential form.** From the first equation: $$xy = 2^3 = 8$$ From the second equation: $$x = 7^2 = 49$$ 4. **Step 3: Substitute $x$ into the first equation to find $y$.** $$49 \times y = 8$$ Divide both sides by 49: $$y = \frac{8}{49}$$ Intermediate step showing cancellation: $$y = \frac{\cancel{8}}{\cancel{49}}$$ (Here, no common factors to cancel, so fraction remains $\frac{8}{49}$.) 5. **Final answer:** $$x = 49, \quad y = \frac{8}{49}$$ This means $x$ is 49 and $y$ is $\frac{8}{49}$, satisfying both logarithmic equations.