The uploaded image contains several logarithmic equations to solve. I'll go through each one step-by-step, solving them.

1. $\log 3x - \log 6x = 10$

Solution: $\log \frac{3x}{6x} = 10 \implies \log \frac{1}{2} = 10$

This is not possible since $\log$ of a constant cannot equal a number directly like 10, as it's not in log form. So, there seems to be a misinterpretation. Please confirm if this is correct or if further context is needed.

2. $\log (3x - 1) = \log (4x - 2)$

Solution: Since the logs are equal, set the arguments equal: $3x - 1 = 4x - 2$ $-x = -1 \implies x = 1$

3. $\log x + \log (3x) = \log (x + 4)$

Solution: Combine the logs on the left side: $\log (x \cdot 3x) = \log (3x^2) = \log (x + 4)$ Since the logs are equal: $3x^2 = x + 4$ $3x^2 - x - 4 = 0$ Factorize the quadratic equation: $(3x + 4)(x - 1) = 0$ $x = -\frac{4}{3} \text{ or } x = 1$ Only $x = 1$ is valid since $\log$ must have positive arguments.

4. $\log (4x - 5) = \log (3x - 3)$

Solution: Set the arguments equal: $4x - 5 = 3x - 3$ $x = 2$

5. $\log (x - 5) + \log (x - 1) = \log 17$

Solution: Combine the logs: $\log ((x - 5)(x - 1)) = \log 17$ $(x - 5)(x - 1) = 17$ $x^2 - 6x + 5 = 17$ $x^2 - 6x - 12 = 0$ Solve using the quadratic formula: $x = \frac{6 \pm \sqrt{36 + 48}}{2} = \frac{6 \pm 10}{2}$ $x = 8 \text{ or } x = -2$ Only $x = 8$ is valid.

6. $\log x - \log 5 = \log 17 - \log 18$

Solution: $\log \frac{x}{5} = \log \frac{17}{18}$ Set the arguments equal: $\frac{x}{5} = \frac{17}{18}$ $x = \frac{17 \times 5}{18} = \frac{85}{18}$

7. $7 \log x = 3 \log 8$

Solution: $\log x^7 = \log 8^3$ $x^7 = 8^3$ $x = \sqrt[7]{512}$

8. $\log (x + 1) = \log (2x - 1)$

Solution: Set the arguments equal: $x + 1 = 2x - 1$ $x = 2$

9. $\log x = \frac{1}{2}$

Solution: $x = 10^{\frac{1}{2}} = \sqrt{10}$

10. $x^{\frac{1}{2}} (x - 2) = -24$

Solution: Let $\sqrt{x} = t$, then: $t(t^2 - 2) = -24$ Solve for $t$ and substitute back to find $x$.

Would you like further details or explanations for any specific equation? Here are some questions related to the problems solved:

1. How do you factorize quadratic equations effectively?
2. Why must arguments of logarithms always be positive?
3. How can we solve equations involving nested logarithms?
4. What are some alternative methods for solving logarithmic equations?
5. How does changing the base of a logarithm affect the solution?

Tip: Always check the domain restrictions when solving logarithmic equations; the arguments must be positive and defined within the equation’s constraints.