# How do I solve this exponential equation?

## I have been trying to use substitution but that doesn't work unless its $4 {x}^{2}$.

Jun 27, 2018

$x = 4 , \mathmr{and} , x = 3$.

#### Explanation:

Let, ${2}^{x} = y . \text{ Then, } {4}^{x} = {\left({2}^{2}\right)}^{x} = {\left({2}^{x}\right)}^{2} = {y}^{2}$.

Also, ${2}^{x + 3} = {2}^{x} \cdot {2}^{3} = 8 y$.

Thus, the eqn. becomes, ${y}^{2} - 3 \cdot 8 y + 128 = 0 ,$

$i . e . , {y}^{2} - 24 y + 128 = 0$.

$\therefore \left(y - 16\right) \left(y - 8\right) = 0$.

$\therefore y = 16 , \mathmr{and} , y = 8$

$\therefore {2}^{x} = 16 = {2}^{4} , \mathmr{and} , y = 8 = {2}^{3}$.

$\therefore x = 4 , \mathmr{and} , x = 3$.