# Solve for x: 16*4^(x+2) - 16*2^(x+1) + 1 = 0?

Nov 19, 2016

$x = - 4$

#### Explanation:

$16 \cdot {4}^{x + 2} - 16 \cdot {2}^{x + 1} + 1 =$
$= {16}^{2} \cdot {4}^{x} - 2 \cdot 16 \cdot {2}^{x} + 1 =$
$= {16}^{2} \cdot {\left({2}^{x}\right)}^{2} - 32 \cdot {2}^{x} + 1 = 0$

Making $y = {2}^{x}$

${16}^{2} {y}^{2} - 32 y + 1 = 0$ solving for $y$ we have

$y = \frac{1}{16} = {2}^{x} = {2}^{- 4}$ then $x = - 4$