1. **State the problems:** We have three problems involving logarithms: - Solve the inequality $$\log_{10}(2x - 5) > \log_{10}(x + 1)$$ - Solve the equation $$2\log_{10}(x) = \log_{10}(50)$$ 2. **Solve the inequality $$\log_{10}(2x - 5) > \log_{10}(x + 1)$$:** - Since $$\log_{10}$$ is an increasing function, the inequality holds if and only if the arguments satisfy: $$2x - 5 > x + 1$$ - Solve for $$x$$: $$2x - 5 > x + 1 \implies 2x - x > 1 + 5 \implies x > 6$$ - Also, the arguments must be positive for the logarithms to be defined: $$2x - 5 > 0 \implies x > \frac{5}{2} = 2.5$$ $$x + 1 > 0 \implies x > -1$$ - Combining domain restrictions and inequality solution: $$x > 6$$ 3. **Solve the equation $$2\log_{10}(x) = \log_{10}(50)$$:** - Use logarithm power rule: $$2\log_{10}(x) = \log_{10}(x^2)$$ - So the equation becomes: $$\log_{10}(x^2) = \log_{10}(50)$$ - Since $$\log_{10}$$ is one-to-one: $$x^2 = 50$$ - Solve for $$x$$: $$x = \pm \sqrt{50} = \pm 5\sqrt{2}$$ - But $$x$$ must be positive because $$\log_{10}(x)$$ is defined only for $$x > 0$$, so: $$x = 5\sqrt{2}$$ **Final answers:** - Inequality solution: $$x > 6$$ - Equation solution: $$x = 5\sqrt{2}$$