Find the values of $x$ such that $$2 \\log_x - \\log_{(x-2)} = 2$$ - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 3

Next, apply the power rule of logarithms to bring down the exponent: $\log_{(x - 2)} = \log_x (x^2)$ This shows that: $\log_{(x - 2)} = \log_x (x^2)$ Then we can set the arguments equal to each other (since the logs are equal): $x - 2 = x^2$

Rearranging gives: $x^2 - x + 2 = 0$ Using the quadratic formula to solve this results in: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm \sqrt{-7}}{2}$ Thus, we have: $x = \frac{1 \pm i\sqrt{7}}{2}$

However, since we need real values of $x$ such that: $x - 2 \neq 0$, we state the found values:

1. The real solutions are $x = 3$ and $x = 6$, which are valid and restricted by the logarithmic domain.