Question

$\frac{1}{2} ^{ x^{2}-3 } =4$

Answer 1

$(\frac{1}{2})^{x^2-3}=4\\ \\ (\frac{1}{2})^{x^2-3}=2^2\\ \\ (\frac{1}{2})^{x^2-3}=(\frac{1}{2})^{-2}\\ \\ x^2-3=-2\\ \\ x^2=1\\ \\ \boxed{x= \pm1}$