Solve the equation: $\;$ $\log_{\sqrt{5}} \left(4^x - 6\right) - \log_{\sqrt{5}} \left(2^x - 2\right) = 2$
Given equation: $\;\;$ $\log_{\sqrt{5}} \left(4^x - 6\right) - \log_{\sqrt{5}} \left(2^x - 2\right) = 2$ $\;\;\; \cdots \; (1)$
i.e. $\;$ $\log_{\sqrt{5}} \left(\dfrac{4^x - 6}{2^x - 2}\right) = 2$
i.e. $\;$ $\dfrac{4^x - 6}{2^x - 2} = \left(\sqrt{5}\right)^2 = 5$
i.e. $\;$ $4^x - 6 = 5 \times 2^x - 10$
i.e. $\;$ $\left(2^2\right)^x - 5 \times 2^x + 4 = 0$
i.e. $\;$ $\left(2^x\right)^2 - 5 \times 2^x + 4 = 0$ $\;\;\; \cdots \; (2)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (3)$
Then equation $(2)$ becomes
$p^2 - 5p + 4 = 0$
i.e. $\;$ $\left(p - 1\right) \left(p - 4\right) = 0$
i.e. $\;$ $p = 1$ $\;$ or $\;$ $p = 4$
Substituting the value of $p$ in equation $(3)$ gives
when $\;$ $p = 1$, $\;$ we have $\;$ $2^x = 1$ $\implies$ $x = \log_2 1 = 0$
when $\;$ $p = 4$, $\;$ we have $\;$ $2^x = 4 = 2^2$ $\implies$ $x = 2$
Substituting $\;$ $x = 0$ $\;$ in the term $\;$ $\log_{\sqrt{5}} \left(4^x - 6\right)$ $\;$ in equation $(1)$ gives
$\log_{\sqrt{5}} \left(4^0 - 6\right) = \log_{\sqrt{5}} \left(1 - 6\right) = \log_{\sqrt{5}} \left(-5\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $x = 0$ $\;$ is not a valid solution.
Substituing $\;$ $x = 2$ $\;$ in equation $(1)$ gives
$\log_{\sqrt{5}} \left(4^2 - 6\right) - \log_{\sqrt{5}} \left(2^2 - 2\right) = 2$
i.e. $\;$ $\log_{\sqrt{5}} \left(\dfrac{10}{2}\right) = 2$
i.e. $\;$ $\log_{\sqrt{5}} 5 = 2$
i.e. $\;$ $\log_{\sqrt{5}} \left(\sqrt{5}\right)^2 = 2$
i.e. $\;$ $2 \log_{\sqrt{5}} \sqrt{5} = 2$
i.e. $\;$ $2 = 2$ $\;\;\;$ which is true.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$