Solve $x - \ln(x) > 1$ without using derivatives

Question

I would like to solve $x - \ln(x) > 1$ without studying the function $f(x) = x - \ln(x) - 1$ and computing its derivative.

I wonder if solving for $x$ in $x - \ln(x) = 1$ is possible.

It could be possible to consider the Lambert $W$-function, since we can rewrite the expression such that we have

\begin{align*} \ln(x) - x &= 1 \\ \ln(xe^{-x}) &= 1 \\ xe^{-x} &= e \\ -xe^{-x} &= -e \end{align*}

Yet it seems impossible and non-elementary to have a "nice" analytical solution to this.

Answer 1

Going on what Martin R said in the comments to the question, using the fact that $\ln(x)$ is defined only for $x>0$ and that $e^x$ is a strictly increasing function,

$x-\ln(x)>1,\ x>0 \implies e^{x-\ln(x)} > e^1,\ x>0 \implies \frac{1}{x}e^x > e,\ x>0 \implies e^x > ex,\ x>0.$

Substituting $x = y+1$ gives:

$e^{y+1} > e(y+1),\ y > -1 \implies e^y> y+1,\ y > -1.$

Various proofs of the final equation for all $y \in \mathbb{R}$ can be found here. In fact, you only need a proof that works for all $y > -1, $ which other people are welcome to do if they so desire (I'm not going to...).

Answer 2

I think you made a mistake in your solution \begin{align*} x-\ln(x) &>1 \\ \ln\left(\frac{e^x}{x}\right) &> 1 \\ \frac{e^x}{x} &= e \\ e^x &= xe \end{align*} So, it has only one solution which is $x=1.$ I hope that helps