If $\log_4 x + \log_2 x = 6$, then $x$ is equal to

### Concept & Formula This problem requires making the bases of the logarithms identical so they can be combined. The change of base formula or the property of base powers is essential here. Base Power Property: $$ \log_{a^n}(x) = \frac{1}{n} \cdot \log_a(x) $$ ### Step-by-Step Solution **Given:** $\log_4(x) + \log_2(x) = 6$ First, convert $\log_4(x)$ to base $2$ so it matches the other term. Notice that $4 = 2^2$. Using the base power property: $\log_{2^2}(x) = \frac{1}{2} \cdot \log_2(x)$ Now substitute this back into the original equation: $\frac{1}{2}\log_2(x) + \log_2(x) = 6$ Factor out $\log_2(x)$: $\log_2(x) \left(\frac{1}{2} + 1\right) = 6$ $\frac{3}{2}\log_2(x) = 6$ Isolate $\log_2(x)$ by multiplying both sides by $\frac{2}{3}$: $\log_2(x) = 6 \cdot \frac{2}{3}$ $\log_2(x) = 4$ Convert the logarithmic equation to its exponential form to solve for $x$: $x = 2^4$ $x = 16$ ### Exam Strategy & Shortcut Instead of algebraic manipulation, you can plug the options into the equation, starting from the easiest numbers. Let's try option (d) $x = 16$: $\log_4(16) + \log_2(16)$ $4^2 = 16$, so the first term is $2$. $2^4 = 16$, so the second term is $4$. $2 + 4 = 6$. This matches the equation perfectly. Option checking is much faster here. ### Common Pitfall A common mistake is attempting to use the product rule directly (e.g., adding the arguments to get $2x$ or multiplying bases) when the bases are different. Logarithmic terms cannot be directly added or combined using the product rule unless their bases are exactly the same. ### Final Answer **Therefore, the correct answer is 16.**