what is log base 2 (x-4)-log base 2 (x-5)=3

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.

Understand how to solve the logarithmic equation log base 2 (x-4) - log base 2 (x-5) = 3. This guide uses properties of logarithms and exponential equations, suitable for Grades 10-12. Learn to apply the quotient rule and rewrite the equation in exponential form to solve for x.

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