How to prove $\ln x<x$? How can we prove that the inequality 
$$\ln~x&lt;x$$

It is trivial in the cases $0&lt;x\le 1$. I couldn't do anything for $x&gt;1$.
This is one way to see it among others.

The function $x\mapsto\ln(1+x)$ is a concave function (it's twice differentiable and its second derivative is strictly negative). Thus it's below all its tangents.

The tangent at the point $(0,0)$ is the line $y=x$. Hence
$$\forall x&gt;0,\, \ln(1+x)\leq x$$

We deduce from this that
$$\forall x&gt;0,\, \ln x&lt;x$$
Well, graphically, you can easily prove that. Plotting $\color {blue}{y = e^x}$ vs $\color{red}{y = x}$ graph, we can see:

$e^x &gt; x$

Take log on both sides (Considering only +ve values of $x$).

$x &gt; \ln x$