$\frac{1}{2}(\log x + \log y)$ will equal $\log \left(\frac{x+y}{2}\right)$ if

### Concept & Formula The problem utilizes the product rule of logarithms and basic algebraic identities to simplify logarithmic equations. $$ \log a + \log b = \log(ab) $$ $$ n \log a = \log(a^n) $$ ### Step-by-Step Solution * **Given:** We need to simplify the left side of the equation using the product rule. $$ \frac{1}{2}(\log x + \log y) = \frac{1}{2}\log(xy) $$ * **Calculation:** Apply the power rule to move the fractional coefficient into the logarithm as an exponent. $$ \frac{1}{2}\log(xy) = \log((xy)^{\frac{1}{2}}) = \log(\sqrt{xy}) $$ * Equate this simplified left side to the right side of the given equation. $$ \log(\sqrt{xy}) = \log\left(\frac{x+y}{2}\right) $$ * Since the bases of the logarithms on both sides are identical (base 10), we can drop the logarithms and equate the arguments. $$ \sqrt{xy} = \frac{x+y}{2} $$ * Square both sides and cross-multiply to solve for the variables algebraically. $$ xy = \frac{(x+y)^2}{4} $$ $$ 4xy = x^2 + 2xy + y^2 $$ * Rearrange the terms to form a quadratic equation. $$ x^2 - 2xy + y^2 = 0 $$ * Factor the perfect square trinomial. $$ (x - y)^2 = 0 $$ * Taking the square root of both sides gives $x - y = 0$, which means $x = y$. ### Exam Strategy & Shortcut You can use the **Option Verification Method** here. If you substitute $x = y$ into the original equation: Left Side: $\frac{1}{2}(\log x + \log x) = \frac{1}{2}(2\log x) = \log x$ Right Side: $\log\left(\frac{x+x}{2}\right) = \log\left(\frac{2x}{2}\right) = \log x$ Since Left Side = Right Side, option (c) is immediately verified as correct without needing to solve the full algebraic expansion. ### Common Pitfall Students often incorrectly distribute the fraction or forget to square the denominator when squaring $\frac{x+y}{2}$, leading to an incorrect polynomial and getting stuck. Always wrap the entire term in parentheses before squaring. ### Final Answer **Therefore, the correct answer is $x = y$.**