Evaluate $\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}$

To solve this expression, we need to apply the laws of logarithms, specifically the power rule: $\log_b(x^n) = n \log_b(x)$ and the definition $\log_b(b^n) = n$.

1. Analyze and Simplify each term

The expression is: $\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}$

• Term 1: $2\log_{3} 9$ Since $9 = 3^2$, we have: $2 \log_{3} (3^2) = 2 \times 2 \log_{3} 3 = 4 \times 1 = \mathbf{4}$

• Term 2: $\log_{3} 81^{-2}$ Since $81 = 3^4$, then $81^{-2} = (3^4)^{-2} = 3^{-8}$. So, $\log_{3} (3^{-8}) = \mathbf{-8}$

• Term 3 (Denominator): $\log_{5} 625$ Since $625 = 5^4$, we have: $\log_{5} (5^4) = \mathbf{4}$

2. Substitute the values back into the expression

Now, substitute these simplified values into the original fraction:

3. Final Calculation

Comparing this result to the given options, we find that it matches option C.

Exam Tip: Whenever you see logarithms with numbers like 9, 81, or 625, always try to express them as powers of the base (e.g., $3^2, 3^4, 5^4$). This makes the $\log_b(b^n) = n$ rule very easy to apply and saves you a lot of time during the exam!