Solve the equation: $\;$ $\dfrac{\log_2 \left(9 - 2^x\right)}{3 - x} = 1$
Given equation: $\;\;$ $\dfrac{\log_2 \left(9 - 2^x\right)}{3 - x} = 1$
i.e. $\;$ $\log_2 \left(9 - 2^x\right) = 3 - x$
i.e. $\;$ $9 - 2^x = 2^{3 - x}$
i.e. $\;$ $9 - 2^x = 2^3 \times 2^{-x}$
i.e. $\;$ $9 \times 2^x - 2^{2x} = 8$
i.e. $\;$ $\left(2^x\right)^2 - 9 \times 2^x + 8 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p^2 - 9p + 8 = 0$
i.e. $\;$ $\left(p - 1\right) \left(p - 8\right) = 0$
i.e. $\;$ $p = 1$ $\;$ or $\;$ $p = 8$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = 1$,
$2^x = 1$ $\implies$ $x = \log_2 1 = 0$
when $\;$ $p = 8$,
$2^x = 8 = 2^3$ $\implies$ $x = 3$
But when $\;$ $x = 3$, $\;$ the Left Hand Side of the given problem becomes
$\dfrac{\log_2 \left(9 - 2^3\right)}{3 - 3} = \dfrac{\log_2 1}{0} = \dfrac{0}{0}$
$\implies$ $x = 3$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$