If $\log \frac{a}{b} + \log \frac{b}{a} = \log (a + b)$, then

### Concept & Formula This problem requires the application of the product rule for logarithms to combine the terms on the left side of the equation. $$ \log x + \log y = \log(xy) $$ Additionally, you must recognize the fundamental logarithm identity that the log of 1 is always 0. $$ \log 1 = 0 $$ ### Step-by-Step Solution * **Given:** An equation involving the sum of two logarithmic fractions. $$ \log \frac{a}{b} + \log \frac{b}{a} = \log(a + b) $$ * **Calculation:** Apply the product rule to combine the two logarithms on the left side into a single logarithm. $$ \log \left( \frac{a}{b} \times \frac{b}{a} \right) = \log(a + b) $$ * Simplify the argument inside the combined logarithm. The numerator and denominator cancel each other out entirely. $$ \log(1) = \log(a + b) $$ * Since the logarithm of 1 to any base is always 0, the left side becomes 0. $$ 0 = \log(a + b) $$ * Assuming the standard base 10 (or any valid base $x$), convert the logarithmic equation into an exponential equation to solve for $a + b$. $$ 10^0 = a + b $$ $$ 1 = a + b $$ ### Exam Strategy & Shortcut Notice immediately that $\frac{a}{b}$ and $\frac{b}{a}$ are reciprocals. When you add their logarithms, you are multiplying reciprocals inside a single log, which always yields $\log(1) = 0$. Therefore, you can jump straight to $0 = \log(a+b)$, which instantly means $a+b = 1$. This mental shortcut saves crucial exam time. ### Common Pitfall A common mistake is incorrectly expanding $\log(a+b)$ on the right side into $\log a + \log b$, which is mathematically invalid. Never confuse the product rule $\log(ab) = \log a + \log b$ with an additive argument $\log(a+b)$. Treat $\log(a+b)$ as a single unbreakable entity until you convert it to exponential form. ### Final Answer **Therefore, the correct answer is $a + b = 1$.**