If $\log_8 x + \log_8 \frac{1}{6} = \frac{1}{3}$, then the value of $x$ is

### Concept & Formula Similar to previous problems, this uses the product rule of logarithms to combine terms, and then evaluates the result using fractional exponents (roots). Product Rule: $$ \log_a(M) + \log_a(N) = \log_a(M \cdot N) $$ Fractional Exponent Rule: $$ a^{1/n} = \sqrt[n]{a} $$ ### Step-by-Step Solution **Given:** $\log_8(x) + \log_8\left(\frac{1}{6}\right) = \frac{1}{3}$ Apply the product rule to combine the terms on the left, as they share the same base ($8$): $\log_8\left(x \cdot \frac{1}{6}\right) = \frac{1}{3}$ $\log_8\left(\frac{x}{6}\right) = \frac{1}{3}$ Convert this logarithmic equation into its exponential equivalent: $8^{1/3} = \frac{x}{6}$ Evaluate the fractional exponent. Raising a number to the power of $\frac{1}{3}$ is the same as taking its cube root: $8^{1/3} = \sqrt[3]{8} = 2$ Substitute this back into the equation: $2 = \frac{x}{6}$ Multiply both sides by $6$ to solve for $x$: $x = 2 \cdot 6$ $x = 12$ ### Exam Strategy & Shortcut Memorize your basic cube roots ($2^3 = 8, 3^3 = 27$, etc.). The moment you see an equation equal to $1/3$ with a base of $8$, immediately recognize that the right side of the exponential equation will resolve to the clean integer $2$. This simplifies the problem mentally to $x/6 = 2$, yielding $12$ almost instantly. ### Common Pitfall Students often struggle with fractional exponents and might mistakenly evaluate $8^{1/3}$ as $8/3$, leading to messy fractions and an incorrect final answer. Always remember that a fractional exponent signifies a root, not division. ### Final Answer **Therefore, the correct answer is 12.**