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

Utilize the change of base formula to simplify the expression:

$log_{1}(4 - x) = log(4 - x)$. Then, we can equate this to our previous expression:

$log(4 - x) = log_{x}(x^2)$.

Based on the properties of logarithms, we can set the arguments equal:

$4 - x = \frac{x^2}{x} = x^2$.

Rearranging this results in:

$x^2 + x - 4 = 0$. This is a quadratic equation that can be solved using the quadratic formula:

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For our equation, $a = 1$, $b = 1$, and $c = -4$, so we have:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$.

Calculating this gives us two potential solutions:

$x = \frac{-1 + \sqrt{17}}{2}$ and $x = \frac{-1 - \sqrt{17}}{2}$. Since $0 < x < 4$, we discard the negative solution and find:

$x = \frac{-1 + \sqrt{17}}{2} \approx 1.561$.