Solve the equation: $\;$ $4^x - 10 \times 2^{x - 1} = 24$
Given equation: $\;\;$ $4^x - 10 \times 2^{x - 1} = 24$ $\;\;\; \cdots \; (1)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$
Then, $\;\;$ $4^x = \left(2^2\right)^x = \left(2^x\right)^2 = p^2$ $\;\;\; \cdots \; (3)$
Equation $(1)$ can be written as
$4^x - 10 \times \dfrac{2^x}{2} = 24$
i.e. $\;$ $4^x - 5 \times 2^x = 24$
i.e. $\;$ $p^2 - 5p - 24 = 0$ $\;\;\;$ [in view of equations $(2)$ and $(3)$]
i.e. $\;$ $\left(p - 8\right) \left(p + 3\right) = 0$
i.e. $\;$ $p = 8$ $\;\;$ or $\;\;$ $p = -3$
When $\;$ $p = 8$, $\;$ we have from equation $(2)$, $\;$ $2^x = 8 = 2^3$ $\implies$ $x = 3$
When $\;$ $p = -3$, $\;$ we have from equation $(2)$, $\;$ $2^x = -3$ $\implies$ $x = \log_2 \left(-3\right)$
But logarithim of a negative number is not defined.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = 3$.