The value of $(\log_9 27 + \log_8 32)$ is

### Concept & Formula This problem requires evaluating two separate logarithms by expressing both their bases and arguments as powers of a common prime number. The fundamental rule to use here is the base-power and argument-power property: $$ \log_{a^k} (b^n) = \frac{n}{k} \cdot \log_a(b) $$ ### Step-by-Step Solution Let's evaluate each logarithmic term individually before adding them. **First term: $\log_9(27)$** Express both $9$ and $27$ as powers of the prime base $3$: $9 = 3^2$ $27 = 3^3$ Rewrite the logarithm: $\log_{3^2}(3^3)$ Apply the power rule to pull out the exponents as a fraction: $\frac{3}{2} \cdot \log_3(3)$ Since $\log_3(3) = 1$, the term simplifies to: $\frac{3}{2}$ **Second term: $\log_8(32)$** Express both $8$ and $32$ as powers of the prime base $2$: $8 = 2^3$ $32 = 2^5$ Rewrite the logarithm: $\log_{2^3}(2^5)$ Apply the power rule: $\frac{5}{3} \cdot \log_2(2)$ Since $\log_2(2) = 1$, the term simplifies to: $\frac{5}{3}$ **Combine the terms:** Now add the simplified values of the two terms together: $\frac{3}{2} + \frac{5}{3}$ To add these fractions, find a common denominator, which is $6$: $\left(\frac{3 \cdot 3}{2 \cdot 3}\right) + \left(\frac{5 \cdot 2}{3 \cdot 2}\right)$ $\frac{9}{6} + \frac{10}{6}$ $\frac{19}{6}$ ### Exam Strategy & Shortcut For any log where the base and argument are powers of the same number, the answer is just the ratio of their exponents: (power of argument) / (power of base). $27$ is $3^3$, $9$ is $3^2 \Rightarrow 3/2$. $32$ is $2^5$, $8$ is $2^3 \Rightarrow 5/3$. You can skip writing down the log notations entirely and jump straight to adding the fractions $3/2 + 5/3$. ### Common Pitfall A very common mistake is flipping the fraction when applying the exponent rule, writing $\log_9(27)$ as $2/3$ instead of $3/2$. Remember the rule: the exponent of the argument goes in the numerator (top), and the exponent of the base goes in the denominator (bottom). ### Final Answer **Therefore, the correct answer is $\frac{19}{6}$.**