# How to solve for x e^(ln2x) = 12?

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#### Explanation

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#### Explanation:

I want someone to double check my answer

3
Mar 8, 2018

See below.

#### Explanation:

I am not sure if this is supposed to be:

$x {e}^{\ln \left(2 x\right)} = 12$ $\boldsymbol{\mathmr{and}}$ ${e}^{\ln \left(2 x\right)} = 12$

I will do this for both, and then you can choose which is correct.

For:

$x {e}^{\ln \left(2 x\right)} = 12$

The law of logarithms state that:

${b}^{{\log}_{b} a} = a$

$\therefore$

$x \left(2 x\right) = 12$

$2 {x}^{2} = 12$

${x}^{2} = \frac{12}{2} = 6$

$x = \pm \sqrt{6}$

Only $\sqrt{6}$ is valid for real numbers:

$\ln \left(x\right)$ requires $x > 0$

So:

$\textcolor{b l u e}{x = \sqrt{6}}$

For:

${e}^{\ln \left(2 x\right)} = 12$

$2 x = 12$

$x = \frac{12}{2} = 6$

$\textcolor{b l u e}{x = 6}$

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