1. **State the problem:** Solve the equation $$\log(5x + 6) = 2 \log_{10}(5x + 6)$$ where the base of the first logarithm is not explicitly given, so we assume it is base 10. 2. **Recall the logarithm properties:** - The logarithm of a number to the same base is equal, so $$\log(5x + 6) = \log_{10}(5x + 6)$$. - The power rule for logarithms: $$a \log_b(c) = \log_b(c^a)$$. 3. **Rewrite the right side using the power rule:** $$2 \log_{10}(5x + 6) = \log_{10}((5x + 6)^2)$$ 4. **Set the equation:** $$\log_{10}(5x + 6) = \log_{10}((5x + 6)^2)$$ 5. **Since the logarithms are equal and have the same base, their arguments must be equal:** $$5x + 6 = (5x + 6)^2$$ 6. **Rewrite the equation:** $$5x + 6 = (5x + 6)^2$$ 7. **Bring all terms to one side:** $$0 = (5x + 6)^2 - (5x + 6)$$ 8. **Factor the right side:** $$0 = (5x + 6)((5x + 6) - 1) = (5x + 6)(5x + 5)$$ 9. **Set each factor equal to zero:** - $$5x + 6 = 0 \Rightarrow x = -\frac{6}{5} = -1.2$$ - $$5x + 5 = 0 \Rightarrow x = -1$$ 10. **Check the domain:** The argument of the logarithm must be positive: - For $$x = -1.2$$, $$5(-1.2) + 6 = -6 + 6 = 0$$ (not positive, so discard) - For $$x = -1$$, $$5(-1) + 6 = -5 + 6 = 1$$ (positive, valid) **Final answer:** $$x = -1$$