The value of $(\log_3 4) (\log_4 5) (\log_5 6) (\log_6 7) (\log_7 8) (\log_8 9)$ is

### Concept & Formula This problem requires the **Logarithmic Chain Rule**, which is derived from the Change of Base formula. It allows you to collapse a product of logarithms when the argument of one logarithm matches the base of the next. $$ \log_a b \times \log_b c = \log_a c $$ ### Step-by-Step Solution * **Given:** We need to find the product of a continuous chain of logarithms. $$ (\log_3 4) \times (\log_4 5) \times (\log_5 6) \times (\log_6 7) \times (\log_7 8) \times (\log_8 9) $$ * **Calculation:** Alternatively to the chain rule, express each logarithm as a fraction using the standard change of base formula to any common base (like base 10). $$ \frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \times \frac{\log 6}{\log 5} \times \frac{\log 7}{\log 6} \times \frac{\log 8}{\log 7} \times \frac{\log 9}{\log 8} $$ * Observe that the numerator of each fraction cancels out with the denominator of the subsequent fraction (e.g., $\log 4$ cancels $\log 4$). $$ \frac{\log 9}{\log 3} $$ * Rewrite this remaining fraction back into a single logarithm. $$ \log_3 9 $$ * Solve the final logarithm by expressing 9 as a power of the base 3. $$ \log_3 (3^2) = 2 \log_3 3 = 2(1) = 2 $$ ### Exam Strategy & Shortcut Use the visual cancellation shortcut. In a product chain like $\log_a b \times \log_b c \times \log_c d$, just slash out the matching diagonals (the $b$'s and $c$'s). You are left with the very first base and the very last argument. For this question: base is 3, final argument is 9. $\log_3 9 = 2$. It takes 5 seconds. ### Common Pitfall Students sometimes confuse logarithmic multiplication with logarithmic addition. Do not mistakenly attempt to add the arguments or bases together. Multiplication of logs only simplifies cleanly when this chain property is applicable. ### Final Answer **Therefore, the correct answer is 2.**