Solve the equation: $\;$ $2^{2+x} - 2^{2-x} = 15$
Given equation: $\;\;$ $2^{2+x} - 2^{2-x} = 15$
i.e. $\;$ $2^2 \times 2^x - 2^2 \times 2^{-x} = 15$
i.e. $\;$ $4 \times 2^x - \dfrac{4}{2^x} - 15 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$4p - \dfrac{4}{p} - 15 = 0$
i.e. $\;$ $4p^2 - 15p - 4 = 0$
i.e. $\;$ $\left(p - 4\right) \left(4p + 1\right) = 0$
i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = \dfrac{-1}{4}$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = 4$, $\;$ $2^x = 4$
i.e. $\;$ $2^x = 2^2$ $\implies$ $x = 2$
when $\;$ $p = \dfrac{-1}{4}$, $\;$ $2^x = \dfrac{-1}{4}$
i.e. $\;$ $2^x = -2^{-2}$ $\implies$ $x = \log_2 \left(-2^2\right)$
But logarithim of a negative number is not defined.
$\therefore \;$ $p = \dfrac{-1}{4}$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2\right\}$