How do you solve 2^(x^2-3)=15?

Oct 18, 2016

$x = \pm 2.628$

Explanation:

${2}^{{x}^{2} - 3} = 15$

$\Leftrightarrow {x}^{2} - 3 = {\log}_{2} 15 = \log \frac{15}{\log} 2 = \frac{1.1761}{0.3010} = 3.9073$

and ${x}^{2} = 3 + 3.9073 = 6.9073$

and $x = \sqrt{6.9073} = \pm 2.628$