Solve Logarithmic Inequality: log₁/₈(2x)/log₁/₈(4) < log₄(5x-2)

What is the domain of X so that the following is satisfied:

$\frac{\log_{\frac{1}{8}}2x}{\log_{\frac{1}{8}}4}<\log_4(5x-2)$

To solve the inequality $\frac{\log_{\frac{1}{8}}(2x)}{\log_{\frac{1}{8}}(4)} < \log_4(5x - 2)$, we proceed as follows:

• Step 1: Convert all logarithms to a common base using the change of base formula:

  $\log_{\frac{1}{8}}(a) = \frac{\log(a)}{\log(\frac{1}{8})}$ and $\log_4(b) = \frac{\log(b)}{\log(4)}$.

• Step 2: Simplify the inequality using these conversions.

  The left expression becomes $\frac{\log(2x)}{\log(\frac{1}{8})} \div \frac{\log(4)}{\log(\frac{1}{8})} = \frac{\log(2x)}{\log(4)}$.

• Step 3: The inequality simplifies to $\frac{\log(2x)}{\log(4)} < \frac{\log(5x - 2)}{\log(4)}$.
• Step 4: Since both sides are divided by the positive $\log(4)$, the inequality remains:

  $\log(2x) < \log(5x - 2)$.

• Step 5: Remove logs since the logarithms are to the same base, leading to $2x < 5x - 2$.
• Step 6: Solve the inequality $2x < 5x - 2$. Rearrange terms: $2 < 3x$.
• Step 7: Divide both sides by 3 to solve for $x$: $\frac{2}{3} < x$.
• Step 8: Validate $(5x - 2) > 0$ implies $x > \frac{2}{5}$, which is consistent with our solution.

Therefore, the solution to the problem is $\frac{2}{3} < x$, which is choice 1.