# How do you solve 2^x^2 = 32(2^(4x))?

Oct 26, 2015

$x = 5 \mathmr{and} - 1$

#### Explanation:

Looking at your format I assume you meant to write:

${2}^{{x}^{2}} = 32 \left({2}^{4 x}\right)$

$\implies \frac{{2}^{{x}^{2}}}{{2}^{4 x}} = 32$

$\implies {2}^{\left({x}^{2} - 4 x\right)} = 32$

Taking logs:

$\implies \left({x}^{2} - 4 x\right) \log \left(2\right) = \log \left(32\right)$

$\implies {x}^{2} - 4 x = \frac{\log \left(32\right)}{\log \left(2\right)}$

This gives us a quadratic of:

${x}^{2} - 4 x - \frac{\log \left(32\right)}{\log \left(2\right)} = 0$

But $\frac{\log \left(32\right)}{\log \left(2\right)} = 5$ giving:

${x}^{2} - 4 x - 5 = 0$

$\left(x - 5\right) \left(x + 1\right) = 0$

so $x = 5 \mathmr{and} - 1$
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Check

$x \left(x + 1\right) - 5 \left(x + 1\right) = {x}^{2} + x - 5 x - 5$

${x}^{2} - 4 x - 5$