Given that 0 < x < 4 and log_{x}(4 - x) - 2log_{x}(x) = 1, find the value of x. - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 2

We simplify the left side: $log_{x}(x^2) = log_{x}(4 - x).$ Thus, we can write: $x^2 = 4 - x.$

Rearranging gives us: $x^2 + x - 4 = 0.$

Using the quadratic formula, where a = 1, b = 1, and c = -4, we have: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$ This results in: $x = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}.$

Calculating further yields two potential values for x: $x = \frac{-1 + \sqrt{17}}{2}$ and $x = \frac{-1 - \sqrt{17}}{2}.$ However, since we must have $0 < x < 4,$ only the solution $x = \frac{-1 + \sqrt{17}}{2}$ is valid.

Thus, the value of x is: $x = \frac{4}{5}.$