Solve Mixed Logarithmic Equation: ln(8x) and log₇(e²) Problem

To solve the problem, we proceed as follows:

Given the equation:

$\ln 8x \times \log_7 e^2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 1: Express $\ln 8x$ using the change of base formula:

• $\ln 8x = \frac{\log_7 (8x)}{\log_7 e}$

• Step 2: Substitute into the original equation:

• $\frac{\log_7 (8x)}{\log_7 e} \cdot \log_7 e^2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 3: Simplify using $\log_7 e^2 = 2 \log_7 e$:

• $\frac{\log_7 (8x)}{\log_7 e} \cdot 2 \log_7 e = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 4: Cancel $\log_7 e$ and simplify:

• $\log_7 (8x) \cdot 2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 5: Cancel 2 on both sides:

• $\log_7 (8x) = \log_7 8 + \log_7 x^2 - \log_7 x$

• Step 6: Use the properties of logarithms:

• $\log_7 (8x) = \log_7 8 + \log_7 \frac{x^2}{x}$

• Step 7: Simplify $\log_7 \frac{x^2}{x}$:

• $\log_7 (8x) = \log_7 8 + \log_7 x$

• Step 8: Use properties $\log_b m + \log_b n = \log_b (mn)$:

• $\log_7 (8x) = \log_7 (8x)$

• Step 9: This equality is true for all $x > 0$, considering domain restrictions:

• $\text{For } x > 0$

Thus, the solution is valid for all $x$ such that $x > 0$

Therefore, the correct solution is, For all $\mathbf{x > 0}$.