Well I don’t have any method to solve it, but if we put anything other than 1, let’s say \(\displaystyle x \gt e \), then we always gonna end up with \(\displaystyle RHS ~\lt~LHS\). We can set up something like this

\(\displaystyle \ln (x)\lt x \) for \(\displaystyle x > 1\)

Since, \(\displaystyle [\ln (x)]\leq \ln(x) \)

Therefore, \(\displaystyle [\ln(x)]\lt x \). Now from the graph of \(\displaystyle x \) and \(\displaystyle [\ln(x)] \) (we can draw it in desmos) we find that difference, that is \(\displaystyle x - [\ln(x)] \) at x > 1 is greater than 1 and it goes on increasing. So, we can write

\(\displaystyle x - [\ln(x)] \gt 1\)

\(\displaystyle x \gt 1 + [\ln(x)] \)

Hence, x = 1 is the only solution.

P.S. : whenever I wrote \(\displaystyle [\ln(x)] \) I meant that greatest integer function.