Let's solve each equation step by step, using logarithmic laws. I'll explain the key steps along the way.

Problem (a):

$\log(15x) = \log(30)$

Solution:

1. Since $\log(15x) = \log(30)$, we can equate the arguments because the logs are equal: $15x = 30$
2. Divide both sides by 15: $x = \frac{30}{15} = 2$
3. Solution: $x = 2$.

Problem (b):

$\log(3x - 2) = 2$

Solution:

1. Rewrite the equation in exponential form (using the fact that if $\log_b(y) = a$, then $y = b^a$): $3x - 2 = 10^2$
2. Simplify $10^2$ to get: $3x - 2 = 100$
3. Add 2 to both sides: $3x = 102$
4. Divide by 3: $x = \frac{102}{3} = 34$
5. Solution: $x = 34$.

Problem (c):

$\log 9 + \log(x - 8) = 4$

Solution:

1. Use the product rule of logarithms: $\log a + \log b = \log(ab)$. $\log(9(x - 8)) = 4$
2. Rewrite in exponential form: $9(x - 8) = 10^4$
3. Calculate $10^4 = 10000$: $9(x - 8) = 10000$
4. Divide by 9: $x - 8 = \frac{10000}{9}$
5. Add 8 to both sides: $x = \frac{10000}{9} + 8$
6. Solution: $x = \frac{10000}{9} + 8$.

Problem (d):

$\log(x + 1) = \log(2x)$

Solution:

1. Since $\log(x + 1) = \log(2x)$, equate the arguments: $x + 1 = 2x$
2. Rearrange the equation: $1 = x$
3. Solution: $x = 1$.

Problem (e):

$\ln(x + 2) - \ln(x - 7) = \ln 6$

Solution:

1. Use the quotient rule of logarithms: $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$. $\ln\left(\frac{x + 2}{x - 7}\right) = \ln 6$
2. Since $\ln\left(\frac{x + 2}{x - 7}\right) = \ln 6$, equate the arguments: $\frac{x + 2}{x - 7} = 6$
3. Cross-multiply: $x + 2 = 6(x - 7)$
4. Expand the right side: $x + 2 = 6x - 42$
5. Rearrange to solve for $x$: $2 + 42 = 6x - x$ $44 = 5x$
6. Divide by 5: $x = \frac{44}{5} = 8.8$
7. Solution: $x = 8.8$.

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Let me know if you need further details on any step!

Related Questions:

1. What are the common logarithmic properties used in problem-solving?
2. How can you verify solutions in logarithmic equations?
3. What happens if there is no solution for a logarithmic equation?
4. What is the difference between logarithmic and exponential equations?
5. Can logarithmic equations have complex solutions?

Tip:

Always check the domain of the variables in logarithmic equations to avoid undefined expressions (e.g., arguments of logs must be positive).