# Solve for x:16\cdot4^{x+2}-16\cdot2^{x+1}+1=0 ?

Solve for $x:16\cdot {4}^{x+2}-16\cdot {2}^{x+1}+1=0$ ?
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habbocowji
$16\cdot {4}^{x+2}-16\cdot {2}^{x+1}+1=$
$={16}^{2}\cdot {4}^{x}-2\cdot 16\cdot {2}^{x}+1$
$={16}^{2}{\left({2}^{x}\right)}^{2}-32\cdot {2}^{x}+1=0$
Making $y={2}^{x}$
${16}^{2}{y}^{2}-32y+1=0$ solving for y we have
$y=\frac{1}{16}={2}^{x}={2}^{-4}$ then $x=-4$

This is what I was looking for. Thanks!