Let's solve each of the logarithmic equations provided in the image.

a. $\log_6 x + \log_6 3 = 2$

Using the logarithm product rule: $\log_b a + \log_b c = \log_b(ac)$, we can combine the logarithms:

$\log_6(3x) = 2$

This means:

$3x = 6^2$

$3x = 36$

$x = \frac{36}{3} = 12$

So, $x = 12$.

c. $\log_2(x + 4) = 1 + \log_2(x - 3)$

Subtract $\log_2(x - 3)$ from both sides:

$\log_2(x + 4) - \log_2(x - 3) = 1$

Using the logarithm quotient rule: $\log_b a - \log_b c = \log_b\left(\frac{a}{c}\right)$:

$\log_2\left(\frac{x + 4}{x - 3}\right) = 1$

This implies:

$\frac{x + 4}{x - 3} = 2^1 = 2$

Cross-multiplying:

$x + 4 = 2(x - 3)$

Expanding and solving for $x$:

$x + 4 = 2x - 6$

$4 + 6 = 2x - x$

$x = 10$

So, $x = 10$.

e. $\log_5(10x + 3) - \log_5(2x - 1) = 2$

Using the logarithm quotient rule again:

$\log_5\left(\frac{10x + 3}{2x - 1}\right) = 2$

This implies:

$\frac{10x + 3}{2x - 1} = 5^2 = 25$

Cross-multiplying:

$10x + 3 = 25(2x - 1)$

Expanding and solving for $x$:

$10x + 3 = 50x - 25$

$3 + 25 = 50x - 10x$

$28 = 40x$

$x = \frac{28}{40} = \frac{7}{10} = 0.7$

So, $x = 0.7$.

Would you like more details on any of these solutions, or do you have any further questions?

Here are five related questions to deepen your understanding:

1. What is the logarithm change of base formula and how can it be applied to solve logarithmic equations?
2. How can you verify the solutions of logarithmic equations by substituting them back into the original equation?
3. How would the equations change if the logarithm bases were different in each problem?
4. What are the restrictions on the domain of the logarithmic functions in these equations?
5. How can logarithmic identities be used to simplify more complex logarithmic expressions?

Tip: When solving logarithmic equations, always check that the solutions fall within the domain of the logarithmic functions involved.