If $\log_{12} 27 = a$, then $\log_6 16$ is

### Concept & Strategy This is an advanced algebraic manipulation problem. The standard strategy is to express everything in terms of a single, simple prime base (like base $2$ or base $3$). By breaking down both the "given" equation and the "target" expression into the same base, you can isolate the core logarithmic variable and substitute it. Change of Base Formula: $$ \log_b(x) = \frac{\log_k(x)}{\log_k(b)} $$ ### Step-by-Step Solution Let's convert everything to base $2$. **Step 1: Simplify the Given Equation** Given: $a = \log_{12}(27)$ Change to base $2$: $a = \frac{\log_2(27)}{\log_2(12)}$ Break down $27$ and $12$ into prime factors: $27 = 3^3$ $12 = 2^2 \times 3$ Substitute these in and expand: $a = \frac{\log_2(3^3)}{\log_2(2^2 \cdot 3)}$ $a = \frac{3 \cdot \log_2(3)}{2 \cdot \log_2(2) + \log_2(3)}$ Since $\log_2(2) = 1$, we get: $a = \frac{3 \cdot \log_2(3)}{2 + \log_2(3)}$ **Step 2: Isolate $\log_2(3)$** Let $x = \log_2(3)$ for easier algebra. $a = \frac{3x}{2 + x}$ Cross-multiply: $a(2 + x) = 3x$ $2a + ax = 3x$ $2a = 3x - ax$ $2a = x(3 - a)$ $x = \frac{2a}{3 - a}$ So, $\log_2(3) = \frac{2a}{3 - a}$. **Step 3: Simplify the Target Expression** We need to find $\log_6(16)$. Change to base $2$: $\log_6(16) = \frac{\log_2(16)}{\log_2(6)}$ We know $16 = 2^4$ and $6 = 2 \times 3$: $= \frac{4 \cdot \log_2(2)}{\log_2(2) + \log_2(3)}$ $= \frac{4}{1 + \log_2(3)}$ **Step 4: Substitute and Solve** Substitute our value of $\log_2(3)$ from Step 2 into the target expression: $= \frac{4}{1 + \left(\frac{2a}{3 - a}\right)}$ Find a common denominator for the bottom: $= \frac{4}{\frac{3 - a + 2a}{3 - a}}$ $= \frac{4}{\frac{3 + a}{3 - a}}$ Multiply by the reciprocal: $= 4 \cdot \frac{3 - a}{3 + a}$ $= \frac{4(3 - a)}{3 + a}$ ### Exam Strategy & Shortcut Problems involving variables and base changes take too long algebraically. Use dummy numbers if you get stuck. Since $a$ is a specific value ($\approx 1.326$), and the options involve $a$, this indicates the algebraic relationship is absolute. However, mastering the algebraic isolation technique shown above is ultimately safer and often just as fast once you get accustomed to swapping out expressions for variables like $x$. ### Common Pitfall A major pitfall is choosing the wrong base for the initial change of base. While base $10$ or base $3$ works, choosing base $2$ is optimal here because $16$ (in the target log) is a clean power of $2$, simplifying the numerator to an integer $4$ instantly. Choosing base $3$ makes the final substitution slightly messier. ### Final Answer **Therefore, the correct answer is $\frac{4 (3 - a)}{(3 + a)}$.**