What is an upper-bound on $x$, given that $x < k \ln(1+x)$? I believe that the solution is something of the form $\mathcal{O}(k \ln k)$ but I am unable to prove this.
This is my first encounter with such an inequality and any pointers to relevant techniques will be helpful. Thanks.
For $k>1$ let $X(k)$ be the unique positive solution of $x=l\log(1+x)$. Then $$ k\,(\log k+\log\log k)<X(k)<k\,\Bigl(\log k+\log\log k+\frac{1}{\sqrt{\log k}}\Bigr). $$ I will prove the first inequality. We want to prove that $$ k\,(\log k+\log\log k)<k\log(1+k\,(\log k+\log\log k)). $$ This is equivalent to $$ \log k+\log\log k<\log(1+k\,(\log k+\log\log k)), $$ which in turn is equivalent to the trivial inequality $$ k\log k<1+k\,(\log k+\log\log k). $$ The other inequality can be proved in a similar way, with a little more work.