Let $a$ and $b$ be real constants where $b$ is positive. What is the small real number $x_0>0$ such that $2^x \ge a+bx?$
Here is what I have tried. Suppose $a=0$. Fix a positive integer $n$. Then $2^x=e^{(\log 2)x}=\left(e^{\frac{\log 2}{n}x}\right)^n\ge \left(1+\frac{\log 2}{n}x\right)^n\gt\left(\frac{\log 2}{n}x\right)^n$. The inequality $\left(\frac{\log 2}{n}x\right)^n\ge bx$ is solved as $x\ge \left[\frac{b}{(\frac{\log 2}{n})^n}\right]^{1/(n-1)}=\frac{b^{1/(n-1)}}{(\log 2)^{n/(n-1)}} n^{n/(n-1)}$. Therefore $x_0\ge \inf_n \frac{b^{1/(n-1)}}{(\log 2)^{n/(n-1)}} n^{n/(n-1)}.$ Can one do better than this or simplify this expression?