Solve the equation: $\;$ $\log_3 \left(9^x + 9\right) = x + \log_3 \left(28 - 2 \times 3^x\right)$
Given equation: $\;\;$ $\log_3 \left(9^x + 9\right) = x + \log_3 \left(28 - 2 \times 3^x\right)$
i.e. $\;$ $\log_3 \left(9^x + 9\right) - \log_3 \left(28 - 2 \times 3^x\right) = x$
i.e. $\;$ $\log_3 \left(\dfrac{9^x + 9}{28 - 2 \times 3^x}\right) = x$
i.e. $\;$ $\dfrac{9^x + 9}{28 - 2 \times 3^x} = 3^x$
i.e. $\;$ $9^x + 9 = 28 \times 3^x - 2 \times 3^{2x}$
i.e. $\;$ $3^{2x} + 9 = 28 \times 3^x - 2 \times 3^{2x}$
i.e. $\;$ $3 \times \left(3^x\right)^2 - 28 \times 3^x + 9 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$
Then, equation $(1)$ becomes
$3p^2 - 28p + 9 = 0$
i.e. $\;$ $\left(3p - 1\right) \left(p - 9\right) = 0$
i.e. $\;$ $p = \dfrac{1}{3}$ $\;$ or $\;$ $p = 9$
Substituting the value of $\;$ $p$ $\;$ in equation $(2)$ gives
when $\;$ $p = \dfrac{1}{3}$, $\;$ we have $\;$ $3^x = \dfrac{1}{3} = 3^{-1}$ $\implies$ $x = -1$
when $\;$ $p = 9$, $\;$ we have $\;$ $3^x = 9 = 3^2$ $\implies$ $x = 2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-1, 2 \right\}$