1. **Problem 2:** Solve for $x$ in the equation $$\ln(x - 2) + \ln(2x - 3) = 2 \ln x.$$ 2. **Formula and rules:** Use the logarithm property $$\ln a + \ln b = \ln(ab)$$ and the power rule $$k \ln a = \ln(a^k).$$ 3. **Apply properties:** $$\ln((x - 2)(2x - 3)) = \ln(x^2).$$ 4. **Set arguments equal:** Since $$\ln A = \ln B \implies A = B,$$ $$(x - 2)(2x - 3) = x^2.$$ 5. **Expand left side:** $$2x^2 - 3x - 4x + 6 = x^2,$$ $$2x^2 - 7x + 6 = x^2.$$ 6. **Bring all terms to one side:** $$2x^2 - 7x + 6 - x^2 = 0,$$ $$x^2 - 7x + 6 = 0.$$ 7. **Factor quadratic:** $$x^2 - 7x + 6 = (x - 6)(x - 1) = 0.$$ 8. **Solve for $x$:** $$x = 6 \text{ or } x = 1.$$ 9. **Check domain restrictions:** - For $\ln(x - 2)$, require $x - 2 > 0 \Rightarrow x > 2$. - For $\ln(2x - 3)$, require $2x - 3 > 0 \Rightarrow x > \frac{3}{2}$. - For $\ln x$, require $x > 0$. Only $x = 6$ satisfies all domain restrictions. --- 1. **Problem 3:** Solve for $x$ in the equation $$\log_2 x + \log_2 (10 - x) = 4.$$ 2. **Formula and rules:** Use $$\log_a b + \log_a c = \log_a (bc).$$ 3. **Combine logs:** $$\log_2 (x(10 - x)) = 4.$$ 4. **Rewrite in exponential form:** $$x(10 - x) = 2^4 = 16.$$ 5. **Expand and form quadratic:** $$10x - x^2 = 16,$$ $$-x^2 + 10x - 16 = 0,$$ $$x^2 - 10x + 16 = 0.$$ 6. **Factor quadratic:** $$x^2 - 10x + 16 = (x - 8)(x - 2) = 0.$$ 7. **Solve for $x$:** $$x = 8 \text{ or } x = 2.$$ 8. **Check domain restrictions:** - $x > 0$ for $\log_2 x$. - $10 - x > 0 \Rightarrow x < 10$ for $\log_2(10 - x)$. Both $x=8$ and $x=2$ satisfy domain restrictions. --- 1. **Problem 4:** Solve for $x$ in the equation $$\log_2 (x + 3) + \log_2 (x - 3) = 4.$$ 2. **Formula and rules:** Use $$\log_a b + \log_a c = \log_a (bc).$$ 3. **Combine logs:** $$\log_2 ((x + 3)(x - 3)) = 4.$$ 4. **Rewrite in exponential form:** $$(x + 3)(x - 3) = 2^4 = 16.$$ 5. **Simplify product:** $$x^2 - 9 = 16.$$ 6. **Solve for $x^2$:** $$x^2 = 16 + 9 = 25.$$ 7. **Take square root:** $$x = \pm 5.$$ 8. **Check domain restrictions:** - $x + 3 > 0 \Rightarrow x > -3$. - $x - 3 > 0 \Rightarrow x > 3$. Only $x = 5$ satisfies both. **Final answers:** - Problem 2: $x = 6$ - Problem 3: $x = 2$ or $x = 8$ - Problem 4: $x = 5$