1. **State the problem:** Solve the equation $\log_x(8x - 12) = 2$ for $x$. 2. **Rewrite the logarithmic equation in exponential form:** Recall that $\log_a b = c$ means $a^c = b$. So here, $x^2 = 8x - 12$. 3. **Bring all terms to one side to form a quadratic equation:** $$x^2 - 8x + 12 = 0$$ 4. **Factor the quadratic:** $$(x - 6)(x - 2) = 0$$ 5. **Solve for $x$:** $$x - 6 = 0 \Rightarrow x = 6$$ $$x - 2 = 0 \Rightarrow x = 2$$ 6. **Check the domain restrictions:** - The base $x$ of the logarithm must be positive and not equal to 1: $x > 0$ and $x \neq 1$. - The argument $8x - 12$ must be positive: $8x - 12 > 0 \Rightarrow x > \frac{12}{8} = 1.5$. 7. **Apply domain restrictions to solutions:** - $x = 6$ is valid since $6 > 1.5$ and $6 \neq 1$. - $x = 2$ is valid since $2 > 1.5$ and $2 \neq 1$. **Final answer:** $$x = 2 \text{ or } x = 6$$