minimum number $\alpha$ such that for all $x$: $\alpha^x \geqslant x$

Question

This is a short question which I'm asking just out of curiosity:

Find the minimum positive number $\alpha \in \mathbb{R}$ such that $\forall x \in \mathbb{R}: \alpha^x \geqslant x$.

Maybe somebody even knows how it relates to other mathematical constants, e.g. to $e$.

You're also welcome with some hints or references to the corresponding literature :).

Answer 1

First, if $x \le 0$, then $\alpha^x > 0 \ge x$ automatically. Otherwise, note that if $x > 0$, then $\alpha^x > x$ is equivalent to $$\ln \alpha > \frac{\ln x}{x}.$$ So, the condition that $\alpha^x > x$ for all $x$ is equivalent to the condition that $\ln \alpha$ is greater than the maximum value of $\frac{\ln x}{x}$ for $x \in (0, \infty)$. From here, a standard calculation using calculus should tell you what that maximum value $M$ is; and then the set of $\alpha$ satisfying your condition will be $(e^M, \infty)$.

(In particular, as stated in the comments, there is no "minimum value" of $\alpha$ making the condition true.)

Answer 2

For tangency condition we need that

• $\alpha^x=e^{x\log \alpha}\implies (\alpha^x)'=\alpha^x\log \alpha=1$
• $\alpha^x=x$

that is

$$\alpha^{\frac1{\log \alpha}}=\frac 1{\log \alpha}\implies\alpha=e^{\frac1e}\approx 1.44466786...$$

thus $\alpha$ is the minimum number such that $\alpha^x \ge x$.

Answer 3

Assume that $\alpha^x>x$ for all $x\in\mathbb R$. Then $\alpha>1$, since otherwise (i.e. for $\alpha\leq 1$) it follows from $\alpha^x>x$ by letting $x\to\infty$ that $1\geq\lim_{x\to\infty}\alpha^x\geq\lim_{x\to\infty}x=\infty$, which is not possible.

The minimum of the function $h(x):=\alpha^x-x$ is attained at $t\in\mathbb R$ which satisfies $h'(t)=0$, that is, $\alpha^t\log(\alpha)=1$, and hence $t=\log(1/\log(\alpha))/\log(\alpha)=-\log(\log(\alpha))/\log(\alpha)$. At this point, the function $h$ has the value $h(t)=\frac{1+\log(\log(\alpha))}{\log(\alpha)}$, which is positive if and only if $\alpha>e^{1/e}$. However, for $\alpha=e^{1/e}$, $h(t)=0$ and hence $\alpha^t=t$. Consequently, such minimal $\alpha$ does not exist.

However, if you are looking for a minimal $\alpha$ such that $\alpha^x\geq x$ for all $x\in\mathbb R$, then $\alpha=e^{1/e}$ is the desired value.

Answer 4

This is only to summarize the answers:

The inequality holds for all $x \leqslant 0$ irrespectively to $\alpha$.

Assuming $x > 0$ we can apply the following transformations: \begin{gather*} \alpha^x > x \\ x\ln{\alpha} > \ln{x} \\ \ln{\alpha} > \frac{\ln{x}}{x} \\ \alpha > e^{\frac{\ln{x}}{x}} \end{gather*} If we want $\alpha$ to be greater than $e^{\frac{\ln{x}}{x}}$ for any $x$ we need to make it greater than maximum of $e^{\frac{\ln{x}}{x}}$. This maximum is some number $M$ and we need to find minimum possible $\alpha$ such that $\alpha > M $.

This is impossible for real numbers $\mathbb{R}$ because for any number $\alpha$ we can find infinitely many numbers $\beta$ such that $M < \beta < \alpha$ - there is no minimum possible number greater than $M$. That is why J.G. was writing that my question was posed incorrectly and that I needed to replace $a^x > x$ with $a^x \geqslant x$.

Let's fix this \begin{gather*} \alpha^x \geqslant x \\ \ldots \\ \alpha \geqslant e^{\frac{\ln{x}}{x}} \end{gather*} Now we can try to find $M$: \begin{gather*} \left(e^{\frac{\ln{x}}{x}}\right)' = 0 \\ \left(\frac{\ln{x}}{x}\right)' = 0 \\ 1 - \ln{x} = 0 \\ x = e; \end{gather*} which gives \begin{equation*} M = e^{\frac{\ln{e}}{e}} = e^{\frac{1}{e}} \end{equation*} The number $\alpha$ which we are seeking is the minimum possible number $\alpha \geqslant M$ - this is indeed $M$.

Therefore \begin{equation*} \alpha = e^{\frac{1}{e}} \end{equation*}