Solve the equation: $\;$ $1 + 2 \log_{\left(x + 2\right)} 5 = \log_5 \left(x + 2\right)$
Given equation: $\;\;$ $1 + 2 \log_{\left(x + 2\right)} 5 = \log_5 \left(x + 2\right)$
i.e. $\;$ $1 + \dfrac{2}{\log_5 \left(x + 2\right)} = \log_5 \left(x + 2\right)$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log_5 \left(x + 2\right) = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$1 + \dfrac{2}{p} = p$
i.e. $\;$ $p^2 - p -2 = 0$
i.e. $\;$ $\left(p - 2\right) \left(p + 1\right) = 0$
i.e. $\;$ $p = 2$ $\;$ or $\;$ $p = -1$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = 2$, $\;$ then $\;$ $\log_5 \left(x + 2\right) = 2$
i.e. $\;$ $x + 2 = 5^2 = 25$ $\implies$ $x = 23$
when $\;$ $p = -1$, $\;$ then $\;$ $\log_5 \left(x + 2\right) = -1$
i.e. $\;$ $x + 2 = 5^{-1} = \dfrac{1}{5}$ $\implies$ $x = \dfrac{-9}{5}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{-9}{5}, \; 23 \right\}$