We are given the logarithmic equation:

$\log\left[(x+3)(x-8)\right] + \log\left(\frac{x+3}{x-8}\right) = 2$

Step 1: Combine the logarithms using logarithmic properties

Using the property $\log a + \log b = \log (a \cdot b)$, we can combine the two logarithms on the left-hand side:

$\log \left( \left[(x+3)(x-8)\right] \cdot \frac{x+3}{x-8} \right) = 2$

Step 2: Simplify the expression inside the logarithm

Simplifying the expression inside the logarithm:

$\left[(x+3)(x-8)\right] \cdot \frac{x+3}{x-8} = \frac{(x+3)(x-8)(x+3)}{(x-8)}$

Cancel out $(x-8)$ from both the numerator and the denominator:

$(x+3)^2$

Thus, the equation becomes:

$\log\left( (x+3)^2 \right) = 2$

Step 3: Remove the logarithm

Since $\log a = b$ means $a = 10^b$, we can rewrite the equation as:

$(x+3)^2 = 10^2 = 100$

Step 4: Solve for $x$

Now solve the equation $(x+3)^2 = 100$. Taking the square root of both sides:

$x+3 = \pm 10$

This gives two possible solutions:

1. $x+3 = 10$, so $x = 7$
2. $x+3 = -10$, so $x = -13$

Step 5: Check for extraneous solutions

We need to ensure that both solutions are valid by checking them in the original logarithmic expressions.

• For $x = 7$:

  • $\log\left((7+3)(7-8)\right) = \log(10 \times (-1))$ involves taking the log of a negative number, which is undefined. Therefore, $x = 7$ is not a valid solution.

• For $x = -13$:

  • $\log\left((-13+3)(-13-8)\right) = \log((-10)(-21)) = \log(210)$, which is valid.

Thus, the only valid solution is $x = -13$.

Final Answer:

$x = -13$

Would you like further details or have any other questions?

Here are 5 follow-up questions related to this topic:

1. What are the general properties of logarithms that allow combining log terms?
2. How do you handle logarithmic equations involving negative arguments?
3. What are some methods to check for extraneous solutions in logarithmic equations?
4. How does the domain of logarithmic functions affect the solution of such equations?
5. Can you solve an equation with natural logarithms like $\ln(x+2) + \ln(x-1) = 1$?

Tip: Always check the domain of a logarithmic function to ensure you don't take the log of a negative number.