Solve for $x$ in the equation $e^x = 2x $

Question

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

How can I solve for x?

Answer 1

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.

Answer 2

$$e^x = 2x$$ $$\text {Take ln of both sides}$$ $$\ln(e^x) = \ln(2x)$$ $$x = \ln(2x)$$

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