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All I can see so far is: $\ln{e^x} = \ln{2x}$ -> $ x = ln{2x} $

How can I solve for x?

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    Are you looking for real $x$ or complex $x$? – 2017-02-27
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    https://en.wikipedia.org/wiki/Lambert_W_function#Examples – 2017-02-27
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    @S.C.B. I do not think this equation has real solutions. Graphing $e^x-2x$ at https://www.desmos.com/calculator/d8oueevfh7 shows that it never crosses the $x$-axis, thus the only roots must be complex. – 2017-02-27
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    @MichaelWang Yes, I proved this in my answer. Thus, why I asked this. – 2017-02-27

2 Answers 2

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Note that as we have that $e^a \ge a+1$ for all real $a$, by puting $a=x-1$we have that $$e^{x-1} \ge x$$ For all $x$. Thus, multiplying $e$ on each side gives us that $$e^x \ge ex >2x \implies e^x >2x$$ As $e>2$. Hence, there are no real solutions to $e^x=2x$. However, if you were to include complex solutions, note that our equatio is equvialent to solving $(-x)e^{-x}=-\frac{1}{2}$ $$x=-\mathrm{W}_{n}\left(-\frac{1}{2}\right)$$ Where $\mathrm{W}_{n}$ is the analytic continuation of the product log function.

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    Out of curiosity, where does your first inequality come from? – 2017-02-27
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    @mrnovice I know this inequality from experience. The proof is in the link. – 2017-02-27
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$$e^x = 2x$$ $$\text {Take ln of both sides}$$ $$\ln(e^x) = \ln(2x)$$ $$x = \ln(2x)$$

These two do not intersect, thus there is no solution.

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    I don't think that drawing the graph is a proof exactly.... – 2017-02-27
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    I simplified as much and I could, and then showed that there is no real solution, – 2017-02-27