Question

Solve the following equations.
${0.5}^{x^2}\cdot 2^{2x+2}={64}^{-1}$

Answer 1

${\left(0.5\right)}^{x^2}\cdot 2^{2x+2}={64}^{-1}$

$\Rightarrow {\left(\frac{5}{10}\right)}^{x^2}\cdot 2^{2x+2}=2^{-6}$

$\Rightarrow 2^{-x^2}\cdot 2^{2x+2}=2^{-6}$

$\Rightarrow 2^{-x^2+2x+2}=2^{-6}$

$\Rightarrow -x^2+2x+2=-6$

$\Rightarrow -(x^2-2x-2)=-\left(6\right)$

$x^2-2x-2=6$

$\Rightarrow x^2-2x-2-6=0$

$x^2-2x-8=0$

$x^2-4x+2x-8=0$

$\Rightarrow x(x-4)+2(x-4)=0$

$\Rightarrow (x-4)(x+2)=0$

$x=4,-2$ solved