Show an equation has 2 solutions Show that $2+\ln(x)=x$ has two solutions at $x&gt;0$


that is equivalent to $2+\ln(x)-x=0$ which mean that there are two values of $x$ where $f(x)=0$.

How should I approach this?
One approach would be to note that $f(x)=2+\mathrm{ln}(x)-x$ is continuous and note that:
$$f(\frac{1}{e^2})=2-2-\frac{1}{e^2}=-\frac{1}{e^2}&lt;0$$
$$f(1)=2+0-1=1&gt;0$$
$$f(e^2)=2+2-e^2&lt;0$$
such that $f$ should have (at least) two zeros.

Note that this approach only shows existence of two solutions, not what they are.
Why bother messing around with analysis at all? Let's solve for $x$!

$ 2+\ln(x)=x$

$ -2 = -x+\ln(x) $

$e^{-2} = e^{-x+\ln(x)}$

$e^{-2} = e^{-x}e^{\ln(x)}$

$-e^{-2} = -xe^{-x} $

Notice here that on the right-hand side, we have a equation of the form $y=ae^a$, where $a&lt;0$. 

To solve for $a$ (and ultimately, x), we can use the Lambert W-function, which is defined as the inverse of the relation y. (I.e. $W(ae^a)=a$)

However for $a&lt;0$ the Lambert W-Function is in fact bijective, and precisely bijective. The two solutions: 

$W(-e^{-2})=-x$

and

$W_{-1}(-e^{-2})=-x$

So therefore: 

$x=-W(-e^{-2})$

OR

$x=-W_{-1} (-e^{-2})$

Addendum: 
Technically, if $-x ∈ (-1,0)$, then we must use the function $W_0(x)$ instead of the function $W_{-1}(x)$. But nonetheless there are two solutions because of the property of the Lambert W-function. It really isn't any different to solving an equation to the form $x^2=16$ and realizing that there must be two solutions, $x=-4, 4$.