We are given the logarithmic equation:

$\log_7(x + 5) - \log_7(x - 1) = 1$

Step 1: Use the logarithm subtraction rule.

The logarithm subtraction rule states that:

$\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$

Apply this to the given equation:

$\log_7\left(\frac{x + 5}{x - 1}\right) = 1$

Step 2: Rewrite the equation in exponential form.

Since $\log_b(A) = C$ implies $A = b^C$, we can rewrite the equation as:

$\frac{x + 5}{x - 1} = 7^1$

Simplifying:

$\frac{x + 5}{x - 1} = 7$

Step 3: Solve for $x$.

Multiply both sides by $x - 1$ to clear the fraction:

$x + 5 = 7(x - 1)$

Expand and solve for $x$:

$x + 5 = 7x - 7$

Subtract $x$ from both sides:

$5 = 6x - 7$

Add 7 to both sides:

$12 = 6x$

Divide by 6:

$x = 2$

Step 4: Verify the solution.

Substitute $x = 2$ back into the original logarithmic expressions to ensure they are valid:

$\log_7(2 + 5) - \log_7(2 - 1) = \log_7(7) - \log_7(1)$

This simplifies to:

$\log_7(7) - \log_7(1) = 1 - 0 = 1$

The solution is valid.

Final Answer:

The solution to the equation $\log_7(x + 5) - \log_7(x - 1) = 1$ is:

$x = 2$

Would you like more details or have any questions? Here are some related problems to practice:

1. Solve $\log_2(x + 3) - \log_2(x - 1) = 2$.
2. Solve $\log_5(3x + 1) = 2$.
3. Solve $\log_{10}(x + 4) = 2 + \log_{10}(x)$.
4. Solve $\log_4(x^2 - 4) = 1$.
5. Solve $2\log_3(x) = \log_3(9)$.

Tip: Always check the domain of the logarithmic expressions to ensure the argument inside the logarithm is positive.