$(\log_5 5) (\log_4 9) (\log_3 2)$ is equal to

### Concept & Formula This problem involves simplifying individual logarithmic terms using the base-power property and then combining them using the reciprocal/chain rules of logarithms. Base-Power Property: $$ \log_{a^n}(b^n) = \log_a(b) $$ Reciprocal Property: $$ \log_a(b) \cdot \log_b(a) = 1 $$ ### Step-by-Step Solution **Given expression:** $(\log_5 5) \cdot (\log_4 9) \cdot (\log_3 2)$ Let's simplify each term one by one. First term: $\log_5(5)$ The logarithm of any base to itself is always $1$. $\log_5(5) = 1$ Second term: $\log_4(9)$ Recognize that both the base and the argument are perfect squares. Express them as powers: $4 = 2^2$ $9 = 3^2$ Substitute these into the term: $\log_{2^2}(3^2)$ Apply the base-power property ($\frac{2}{2} = 1$), reducing the term to: $\log_2(3)$ Third term remains as is: $\log_3(2)$ Now, bring all the simplified terms back together into the original multiplication expression: $1 \cdot (\log_2 3) \cdot (\log_3 2)$ $(\log_2 3) \cdot (\log_3 2)$ Use the reciprocal property of logarithms. Since the base of one is the argument of the other, their product is $1$: $\log_2(3) \times \frac{1}{\log_2(3)} = 1$ ### Exam Strategy & Shortcut Whenever you see a logarithm where both the base and argument are raised to the same power, strip the powers immediately. $\log_4(9)$ is instantly just $\log_2(3)$. From there, you have a chain of $\log_2(3) \times \log_3(2)$. The bases and arguments criss-cross and cancel perfectly, leaving $1$. The $\log_5(5)$ is just a distractor that also equals $1$. $1 \times 1 = 1$. ### Common Pitfall A common trap is trying to convert everything into base $10$ fractions right away without simplifying the perfect squares first. While $\frac{\log 9}{\log 4} \times \frac{\log 2}{\log 3}$ will eventually yield the right answer, it creates unnecessary and messy arithmetic steps that consume valuable exam time. ### Final Answer **Therefore, the correct answer is 1.**