Solve the equation: $\;$ $3^{2x+5} = 3^{x+2} + 2$
Given equation: $\;\;$ $3^{2x+5} = 3^{x+2} + 2$
i.e. $\;$ $3^{2x} \times 3^5 - 3^x \times 3^2 - 2 = 0$
i.e. $\;$ $243 \times \left(3^x\right)^2 - 9 \times 3^x - 2 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$243 p^2 - 9p -2 = 0$
i.e. $\;$ $\left(27p + 2\right) \left(9p - 1\right) = 0$
i.e. $\;$ $p = \dfrac{-2}{27}$ $\;$ or $\;$ $p = \dfrac{1}{9}$
Substituing the value of $p$ in equation $(2)$ gives
when $\;$ $p = \dfrac{-2}{27}$, $\;$ then $\;$ $3^x = \dfrac{-2}{27}$
$\implies$ $x = \log_3 \left(\dfrac{-2}{27}\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $p = \dfrac{-2}{27}$ $\;$ is not a valid solution.
when $\;$ $p = \dfrac{1}{9}$, $\;$ then $\;$ $3^x = \dfrac{1}{9} = 3^{-2}$ $\implies$ $x = -2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-2 \right\}$