1. **State the problem:** Solve the equation $$^m\log(2x + 6) \cdot ^5\log m = 2$$ for $x$. 2. **Recall the change of base formula:** For any positive numbers $a,b,c$ with $a \neq 1$, $$^a\log b = \frac{\log b}{\log a}$$ where $\log$ is the logarithm in any fixed base (e.g., base 10 or $e$). 3. **Rewrite the given equation using the change of base formula:** $$^m\log(2x + 6) = \frac{\log(2x + 6)}{\log m}$$ $$^5\log m = \frac{\log m}{\log 5}$$ 4. **Substitute these into the equation:** $$\frac{\log(2x + 6)}{\log m} \cdot \frac{\log m}{\log 5} = 2$$ 5. **Simplify by canceling $\log m$:** $$\frac{\cancel{\log m} \cdot \log(2x + 6)}{\cancel{\log m} \cdot \log 5} = 2$$ which simplifies to $$\frac{\log(2x + 6)}{\log 5} = 2$$ 6. **Multiply both sides by $\log 5$:** $$\log(2x + 6) = 2 \log 5$$ 7. **Use the logarithm power rule:** $$2 \log 5 = \log 5^2 = \log 25$$ 8. **Equate the arguments of the logarithms:** Since $\log(2x + 6) = \log 25$, we have $$2x + 6 = 25$$ 9. **Solve for $x$:** $$2x = 25 - 6 = 19$$ $$x = \frac{19}{2} = 9.5$$ 10. **Check domain restrictions:** The argument of the logarithm must be positive: $$2x + 6 > 0 \implies x > -3$$ Since $9.5 > -3$, the solution is valid. **Final answer:** $$x = 9.5$$