To solve the equation:

$\log_2 (x - 4) - \log_2 (x - 5) = 3$

we can use the following properties of logarithms:

1. Quotient Rule: $\log_b (A) - \log_b (B) = \log_b \left(\frac{A}{B}\right)$
2. Exponent Rule: If $\log_b (A) = C$, then $A = b^C$

Step 1: Apply the Quotient Rule

Using the quotient rule, we rewrite the left side of the equation:

$\log_2 \left(\frac{x - 4}{x - 5}\right) = 3$

Step 2: Rewrite in Exponential Form

Now, we can rewrite this logarithmic equation in exponential form. Since $\log_2 \left(\frac{x - 4}{x - 5}\right) = 3$, we have:

$\frac{x - 4}{x - 5} = 2^3$

Calculating $2^3$, we get:

$\frac{x - 4}{x - 5} = 8$

Step 3: Solve for $x$

Now, we can solve for $x$ by clearing the fraction. Multiply both sides by $x - 5$:

$x - 4 = 8(x - 5)$

Expanding the right side:

$x - 4 = 8x - 40$

Next, isolate $x$ by moving terms involving $x$ to one side:

$-4 + 40 = 8x - x$

$36 = 7x$

Now, divide by 7:

$x = \frac{36}{7}$

Step 4: Verify the Solution

To ensure this solution is valid, we check that $x = \frac{36}{7}$ does not make any arguments of the logarithms negative:

1. $x - 4 = \frac{36}{7} - 4 = \frac{36 - 28}{7} = \frac{8}{7}$, which is positive.
2. $x - 5 = \frac{36}{7} - 5 = \frac{36 - 35}{7} = \frac{1}{7}$, which is also positive.

Thus, $x = \frac{36}{7}$ is a valid solution.

Final Answer

$x = \frac{36}{7}$

Would you like more details on any part?

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Here are 5 related questions you could explore next:

1. How would this process differ for other bases, like $\log_3$ instead of $\log_2$?
2. What if the equation was $\log_2(x - 4) + \log_2(x - 5) = 3$?
3. Can logarithmic equations have no solution? If so, under what conditions?
4. How can you verify solutions for logarithmic equations more systematically?
5. What happens if there’s a coefficient in front of the logarithmic terms?

Tip: Always verify that the solution satisfies the domain restrictions, as logarithmic arguments must be positive.