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

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

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Answer 1 · 2016-10-18T12:49:46

$x = \pm 2.628$ $x=+-2.628$

Explanation:

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

$\Leftrightarrow x^{2} - 3 = {log}_{2} 15 = \frac{log 15}{log 2} = \frac{1.1761}{0.3010} = 3.9073$ $hArrx^2-3=log_2 15=log15/log2=1.1761/0.3010=3.9073$

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

and $x = \sqrt{6.9073} = \pm 2.628$ $x=sqrt6.9073=+-2.628$

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

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