Solve the equation: $\;$ $5^x - 24 = \dfrac{25}{5^x}$
Given equation: $\;\;$ $5^x - 24 = \dfrac{25}{5^x}$ $\;\;\; \cdots \; (1)$
Let $\;$ $5^x = p$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes
$p^2 - 24p - 25 = 0$
i.e. $\;$ $\left(p - 25\right) \left(p + 1\right) = 0$
i.e. $\;$ $p = 25$ $\;\;$ or $\;\;$ $p = -1$
When $\;$ $p = 25$, $\;$ we have from equation $(2)$,
$5^x = 25 = 5^2$ $\implies$ $x = 2$
When $\;$ $p = -1$, $\;$ we have from equation $(2)$,
$5^x = -1$ $\implies$ $x = \log_5 \left(-1\right)$
But logarithom of a negative number is not defined.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = 2$.