Given that $a > b > 0$ and that $a$ and $b$ satisfy the equation $$ ext{log } a - ext{log } b = ext{log }(a - b) $$ (a) show that $$ a = \frac{b^2}{b - 1} $$ (b) Write down the full restriction on the value of $b$, explaining the reason for this restriction. - Edexcel - A-Level Maths Pure - Question 11 - 2019 - Paper 2

To show that $a = \frac{b^2}{b - 1}$, we start from the given equation:

$ext{log } a - ext{log } b = ext{log }(a - b).$

Using the properties of logarithms, this can be rewritten as:

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

Since both sides are equal in log scale, we can equate the arguments:

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

Then we multiply both sides by $b$:

$a = b(a - b).$

Expanding this, we have:

$a = ab - b^2.$

Now, rearranging gives us:

$a - ab + b^2 = 0$

This can be factored as:

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

Solving for $a$, we get:

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

Since we need to express $a$ in terms of $b$, divide both sides by $(1 - b)$, resulting in:

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