If $\log_a m = x$, then $\log_{1/a} \left(\frac{1}{m}\right)$ equals

### Concept & Formula This problem deals with the manipulation of logarithmic bases and arguments when both are expressed as reciprocals (or negative powers). The relevant property for logarithms with exponents in both the base and argument is: $$ \log_{a^k} (b^n) = \frac{n}{k} \cdot \log_a (b) $$ ### Step-by-Step Solution **Given:** $\log_a (m) = x$ We need to find the value of $\log_{1/a} \left(\frac{1}{m}\right)$. First, rewrite both the base and the argument as powers of their original variables. The reciprocal of a number is simply that number raised to the power of $-1$: $\frac{1}{a} = a^{-1}$ $\frac{1}{m} = m^{-1}$ Substitute these into the target expression: $\log_{a^{-1}} (m^{-1})$ Now, apply the exponent property of logarithms, bringing the exponent of the argument to the numerator, and the exponent of the base to the denominator: $$ \log_{a^{-1}} (m^{-1}) = \frac{-1}{-1} \cdot \log_a (m) $$ Simplify the fraction: $$ 1 \cdot \log_a (m) = \log_a (m) $$ Since we were given that $\log_a (m) = x$, substitute $x$ back into the simplified expression: $$ \log_a (m) = x $$ ### Exam Strategy & Shortcut A general rule of thumb for logarithms: if you invert BOTH the base and the argument, the value of the logarithm remains completely unchanged. Since both $a$ and $m$ are inverted to $1/a$ and $1/m$, the negative signs cancel out instantly. You can immediately select $x$ without writing down a single step. ### Common Pitfall The most frequent mistake is pulling out a single negative sign from either the base or the argument, leading students to incorrectly choose $-x$. Remember that both the base and argument contribute a negative sign to the coefficient, which multiply to become positive. ### Final Answer **Therefore, the correct answer is $x$.**