1. **Problem 1:** Solve for $x$ in the equation $$\log_5(10x) - 1 = \log_5(3x - 1)$$ 2. **Problem 2:** Solve for $x$ in the equation $$\log_3(8x - 15) - 1 = 4$$ --- ### Problem 1 Steps: 1. Recall the logarithm property: $$\log_b(a) - c = \log_b\left(\frac{a}{b^c}\right)$$. 2. Rewrite the left side: $$\log_5(10x) - 1 = \log_5\left(\frac{10x}{5^1}\right) = \log_5(2x)$$ 3. So the equation becomes: $$\log_5(2x) = \log_5(3x - 1)$$ 4. Since the logs are equal and the base is the same, their arguments must be equal: $$2x = 3x - 1$$ 5. Solve for $x$: $$2x - 3x = -1 \implies -x = -1 \implies x = 1$$ 6. Check domain restrictions: - Arguments inside logs must be positive: $$10x > 0 \Rightarrow x > 0$$ $$3x - 1 > 0 \Rightarrow x > \frac{1}{3}$$ Since $x=1$ satisfies both, it is valid. --- ### Problem 2 Steps: 1. Start with the equation: $$\log_3(8x - 15) - 1 = 4$$ 2. Use the property: $$\log_3(8x - 15) - 1 = \log_3\left(\frac{8x - 15}{3^1}\right) = \log_3\left(\frac{8x - 15}{3}\right)$$ 3. So the equation becomes: $$\log_3\left(\frac{8x - 15}{3}\right) = 4$$ 4. Rewrite the logarithmic equation in exponential form: $$\frac{8x - 15}{3} = 3^4 = 81$$ 5. Multiply both sides by 3: $$8x - 15 = 243$$ 6. Solve for $x$: $$8x = 243 + 15 = 258$$ $$x = \frac{258}{8} = 32.25$$ 7. Check domain restrictions: $$8x - 15 > 0 \Rightarrow 8(32.25) - 15 = 258 - 15 = 243 > 0$$ Valid solution. --- ### Final answers: - Problem 1: $$x = 1$$ - Problem 2: $$x = 32.25$$