My friend came across this strange equation and I cant find mathematical way to find $x$ without drawing $x$ and $-\ln(x)$ and see that they come across at almost $x=0.5$.
Can any one help?
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$\begingroup$$$x+\ln x=0$$ $$e^{x+\ln x}=e^0=1$$ $$xe^x=1$$
This can be solved by lambert $W$:
$$x=W(1)$$
There is a special name to this constant, it is called the omega constant.
You can find the numerical approximation via Newtons method.
Let $x_1=0.5$
And
$$x_{n+1}=x_{n}-\frac{x_n+\ln x_n}{1+\frac{1}{x_n}}$$
As
$$n \to \infty$$
Then you get closer to the approximate value:
$$x=.56714329...$$
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