If $\log x + \log y = \log (x + y)$, then

### Concept & Formula This question requires combining logarithms using the product rule, and then using basic algebraic manipulation to isolate one variable in terms of the other. $$ \log a + \log b = \log(ab) $$ ### Step-by-Step Solution * **Given:** The equality mapping the sum of logs to the log of a sum. $$ \log x + \log y = \log(x + y) $$ * **Calculation:** Apply the product rule to the left side of the equation. $$ \log(xy) = \log(x + y) $$ * Since the logarithm function is one-to-one, and the bases are the same (base 10 implicitly), we can equate the arguments directly. $$ xy = x + y $$ * The options require $y$ to be expressed in terms of $x$. Bring all terms containing $y$ to one side of the equation. $$ xy - y = x $$ * Factor out $y$ on the left side to isolate it. $$ y(x - 1) = x $$ * Divide both sides by $(x - 1)$ to solve for $y$. $$ y = \frac{x}{x - 1} $$ ### Exam Strategy & Shortcut While the algebra here is straightforward, you can also use substitution to double-check. Pick a value for $x$, say $x = 2$. $xy = x + y \Rightarrow 2y = 2 + y \Rightarrow y = 2$. Now test the options with $x = 2$: (c) $y = \frac{2-1}{2} = 0.5$ (Incorrect) (d) $y = \frac{2}{2-1} = \frac{2}{1} = 2$ (Correct) This confirms the algebraic derivation perfectly. ### Common Pitfall A widespread misconception is that $\log(x + y) = \log x + \log y$ is a general rule. It is absolutely false in standard mathematics. The problem states it as a *conditional* specific scenario to solve for a relationship, not a mathematical identity. Treat it merely as a given equation. ### Final Answer **Therefore, the correct answer is $y = \frac{x}{x-1}$.**