# Does the smallest real number that satisfies $2^x\ge bx$ have logarithmic order?

Thank you for the answers to my earlier question. By working through all your answers, I arrive at a more refined question below.

Fix a positive real constant $b$. Let $x_0$ be the smallest nonnegative real number such that the inequality $2^x\ge bx$ holds for all $x\ge x_0$. Does there exist a constant $C>0$ such that $x_0≤C \log_2 b$?

I showed that $x_0$ has order $O(b^\epsilon)$ in my workings in the earlier question.

-

For every $x\geqslant4$, $2^{x/2}\geqslant x$. Hence, if $x\geqslant4$ and $2^{x/2}\geqslant b$, then $2^x=2^{x/2}\cdot 2^{x/2}\geqslant bx$. To sum up:
If $x\geqslant\max\{4,2\log_2b\}$, then $2^x\geqslant bx$.
In particular, for every $b\geqslant4$, if $x\geqslant2\log_2b$, then $2^x\geqslant bx$, that is, the assertion you are interested in holds with $C=2$. (But note that $x_0(b)\sim\log_2b$ when $b\to+\infty$.)