# How to solve the inequality $2^x\ge a+bx$?

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?

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You're talking about a transcendental equation, so that the solutions (if any) can be determined only if the parameters $a, b$ take particular values - that is, when the equation is a "nice" one and leads to some "trick" for expressing its roots in terms of elementary functions. Other than that, you can only use numerical methods to find arbitrary-precision approximations.

Addendum: of course, the answer using Lambert's W function, although simplifying matters, does not change anything at all: just define the $\diamond$-function to be

$\diamond(a) := F^{-1}(a)$, where $F(t) = 2^t - t$,

and have fun writing a $\mathtt{Mathematica}$ package. Not offending anyone, the point of this is that the solution to $2^x = a + bx$ is not expressible through elementary functions, since the OP seemed to be wanting an explicit $x_0$.

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We can write, for $b\gt0$, \begin{align} e^{x\log2} &\geq a+bx\\ -\frac{1}{b}(\log2)e^{x\log2}&\leq -\frac{a}{b}\log2-x\log2\\ -\frac{1}{b}(\log2)e^{-\frac{a}{b}\log2}e^{\frac{a}{b}\log2+x\log2}&\leq -\frac{a}{b}\log2-x\log2 \end{align} Set $$w=-\frac{a}{b}\log2-x\log2\\ z_0=-\frac{1}{b}(\log2)e^{-\frac{a}{b}\log2}$$ so that we can write $$we^w\geq z_0$$ This can be solved using Lambert $W$ function, defined as the inverse function of $we^w$. This inverse is not unique. Usually the inverse in $[-1,+\infty)$ is called the principal branch and indicated $W_0$. The inverse in $(-\infty,-1]$ is indicated $W_{-1}$.

If $z_0\leq-1/e$ then all $w$ (and so all $x$) are solutions.

If $z_0>-1/e$ then the solutions are $$w\leq W_{-1}(z_0)\qquad\text{or}\qquad w\geq W_0(z_0)$$ These values can be easily evaluated numerically, for example in Mathematica $W_0$ is ProductLog[z] and $W_{-1}$ is ProductLog[-1,z]. The endpoints of the intervals could be easily written in terms of $x$, given the linear relation between $w$ and $x$.

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First note that such $x_0$ can not exists. In fact, $2^x > -1+x, \ \forall x$.Note too that $2^x$ is convex. In fact,$(2^x)´´=(e^{x\ln 2})´´=´(\ln 2)^{2}e^{x\ln 2}\ge0$. Then if there exist $x>0$ such that $a+bx\ge 2^x$ the small $x_0$ such that $2^x\ge a+bx$ will be the small root of the equation $2^x=a+bx$.

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You write $2^{x}=e^{2\ln x}$ but it should be $e^{x \ln 2}$ – Daniel Littlewood Aug 5 '12 at 17:35