We are given that $a > b > 0$ and the equation:

$\log a - \log b = \log(a - b)$

We need to show that:

$a = \frac{b^2}{b - 1}$

Step 1: Simplify the logarithmic equation

Using the logarithmic property $\log x - \log y = \log \left( \frac{x}{y} \right)$, we can simplify the left-hand side:

$\log \left( \frac{a}{b} \right) = \log(a - b)$

Step 2: Exponentiate both sides

Since the logarithms are equal, we can remove the logarithms by exponentiating both sides:

$\frac{a}{b} = a - b$

Step 3: Solve for $a$

Multiply both sides by $b$ to eliminate the denominator:

$a = b(a - b)$

Distribute $b$ on the right-hand side:

$a = ab - b^2$

Now, isolate $a$ on one side by subtracting $ab$ from both sides:

$a - ab = -b^2$

Factor out $a$ on the left-hand side:

$a(1 - b) = -b^2$

Finally, solve for $a$ by dividing both sides by $1 - b$:

$a = \frac{-b^2}{1 - b}$

Since $1 - b$ is negative, this simplifies to:

$a = \frac{b^2}{b - 1}$

Thus, we have shown that:

$a = \frac{b^2}{b - 1}$

Would you like any further clarification or more details on this?

Here are some related questions to consider:

1. How does the property of logarithms simplify this equation?
2. What happens if we don't assume $a > b > 0$?
3. Can this result be applied to other forms of logarithmic equations?
4. How would this result change if we used natural logarithms instead of base-10 logarithms?
5. What are the conditions on $b$ to ensure $a$ remains positive?

Tip: When dealing with logarithmic equations, always check whether using properties like $\log x - \log y = \log \frac{x}{y}$ simplifies the equation early on.