# Hello friends, please, I need to know how to get the answer out of this exponential equation? thank you

Nov 5, 2016

No real solutions

#### Explanation:

${2}^{4 x} - {16}^{x + 3} - {4}^{x - 1} = 0 \to {4}^{2 x} - {4}^{2 \left(x + 3\right)} - {4}^{x} / 4 = 0$ or
$4 \cdot {4}^{2 x} - {4}^{7} \cdot {4}^{2 x} - {4}^{x} = 0$ or
$\left(\left(4 - {4}^{7}\right) {4}^{x} - 1\right) {4}^{x} = 0$

but ${4}^{x} \ne 0$ so

$\left(4 - {4}^{7}\right) {4}^{x} - 1 = 0$ then

${4}^{x} = - \frac{1}{4 \left({4}^{6} - 1\right)}$

but for $x \in \mathbb{R}$ we have ${4}^{x} > 0$ so no real solutions