Let's analyze and solve the logarithmic equations shown in the image:

Equation (a):

$\log_5(2x) = 3$

1. Rewrite the logarithmic equation in exponential form: $2x = 5^3$ $2x = 125$
2. Solve for $x$: $x = \frac{125}{2} = 62.5$

Solution: $x = 62.5$

Equation (b):

$\log_5(x+1) = -1$

1. Rewrite the logarithmic equation in exponential form: $x+1 = 5^{-1}$ $x+1 = \frac{1}{5}$
2. Solve for $x$: $x = \frac{1}{5} - 1 = \frac{-4}{5}$

Solution: $x = -\frac{4}{5}$

Equation (c):

$\log(4) - \log(x) = 2$

1. Use the logarithmic property $\log(a) - \log(b) = \log\left(\frac{a}{b}\right)$: $\log\left(\frac{4}{x}\right) = 2$
2. Rewrite the logarithmic equation in exponential form: $\frac{4}{x} = 10^2$ $\frac{4}{x} = 100$
3. Solve for $x$: $x = \frac{4}{100} = 0.04$

Solution: $x = 0.04$

Equation (d):

$2\log_3(6) + \log_3(y) = 4$

1. Simplify $2\log_3(6)$ using the property $a\log_b(c) = \log_b(c^a)$: $2\log_3(6) = \log_3(6^2) = \log_3(36)$ So the equation becomes: $\log_3(36) + \log_3(y) = 4$
2. Use the logarithmic property $\log_b(a) + \log_b(c) = \log_b(ac)$: $\log_3(36y) = 4$
3. Rewrite the logarithmic equation in exponential form: $36y = 3^4$ $36y = 81$
4. Solve for $y$: $y = \frac{81}{36} = \frac{9}{4} = 2.25$

Solution: $y = 2.25$

Summary of Solutions:

(a) $x = 62.5$
(b) $x = -\frac{4}{5}$
(c) $x = 0.04$
(d) $y = 2.25$

Do you want any of these steps explained in more detail, or have any questions?

Related Questions:

1. What are the basic properties of logarithms used to solve equations?
2. How do you convert a logarithmic equation to its exponential form?
3. Can you explain how to check if a logarithmic solution is valid (e.g., $x > 0$)?
4. How does the base of the logarithm affect the equation's solution?
5. What happens if logarithmic equations have multiple logs on both sides?

Tip:

When solving logarithmic equations, always ensure the arguments of the logarithms are positive, as logarithms are undefined for non-positive values.