Solve the equation: $\;$ $\log_3 \left(3^x - 8\right) = 2 - x$
Given equation: $\;\;$ $\log_3 \left(3^x - 8\right) = 2 - x$
i.e. $\;$ $3^x - 8 = 3^{2 - x}$
i.e. $\;$ $3^x - 8 = 3^2 \times 3^{-x}$
i.e. $\;$ $3^{2x} - 8 \times 3^x - 9 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
i.e. $\;$ $p^2 - 8p - 9 = 0$
i.e. $\;$ $\left(p + 1\right) \left(p - 9\right) = 0$
i.e. $\;$ $p = -1$ $\;$ or $\;$ $p = 9$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = -1$
$3^x = -1$ $\implies$ $x = \log_3 \left(-1\right)$
But logarithim of a negative number is not defined.
$\therefore \;$ $p = -1$ $\;$ is not a valid solution.
when $\;$ $p = 9$
$3^x = 9 = 3^2$ $\implies$ $x = 2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$