# Finding all positive $a$ such that $a^x \geq x$ for all real $x$

Problem:
Given $a\in \mathbb{R}^+$ such that $a^x \geq x$ for all $x\in\mathbb{R}$, what is the value of $a$?

Attempt:
I tried to consider $f(x)=a^x-x$ and its first derivative $f'(x)=a^x \ln a-1$. There are two cases: whether $f'$ has a zero in $\mathbb{R}$ or not.

If $f'$ has no zero, then either $f'(x)>0$ or $f'(x)<0$ for all $x\in\mathbb{R}$. This means that $a^x \ln a >1$ or $a^x \ln a <1$ for all $x\in\mathbb{R}^+\cup{0}$.
I do not know how to proceed from here. I feel that $f'(x)<0$ may not be true but I cannot disprove it. In fact, even for the case $f'(x)>0$, I do not know what to do next.

If $f'(x)$ has a zero at $x=x_0$, then I obtain $x_0 =-\frac{\ln\ln a}{\ln a}$. Plugging $x_0$ into $f(x)$, I obtain $$f(x_0)=a^{-\frac{\ln\ln a}{\ln a}}+\frac{\ln\ln a}{\ln a}$$ This is indeed the minimum point since $f''(x)=a^x (\ln a)^2>0$ for all $x\in\mathbb{R}$ and $a\in \mathbb{R}^+$ (unless $a=1$ which is impossible for the inequality to happen).
What can I deduce from $f(x_0)$ in this case? Is it useful to answer this question?

Would it be easier to be solved if I only look at $x\in \mathbb{R}^+$?

Edited
Indeed, I am looking at all possible values of $a$ such that the inequality is true.
I just noticed (from the answers below) that there is no need to consider $x \leq 0$ since it is $a^x \geq 0$ for $x \leq 0$.

My question has been solved. Thank you for all the response. I really appreciate that.

• "There are two cases: whether $f'$ has a zero in $\Bbb R$ or not." It will always have a zero, assuming $a\neq1$. – Arthur Apr 9 '18 at 17:48
• Note: it's not entirely clear what you are looking for. I am assuming that you meant "for which $a$ is it true that $a^x≥x$ $\forall x\in \mathbb R$, yes? In any case, I posted some calculations relevant to that below. – lulu Apr 9 '18 at 17:49
• Why would you assume there is only one value of $a$? If $b > a$ then $b^x > a^x$ for all $x > 0$. And for $x < 0$ then $b^x > 0 > x$ so there can't be just one. – fleablood Apr 9 '18 at 17:49
• "Would it be easier to be solved if I only look at x∈R+" Well, if $x \le 0$ then $a^x \ge 0 \ge x$ so ... yes. – fleablood Apr 9 '18 at 17:51

There is a range of $a$ that works. It's reasonably clear that if $a$ works then $b>a\implies b$ works as well. Thus we are searching for the least $a$ that works. We note, for instance, that $a=1.01$ is too small while $a=2$ is certainly big enough.

Let $$f_a(x)=a^x-x$$

It is reasonably clear that $f_a$ has a unique minimum. We want that minimum to coincide with the root. If we have a root $x_a$ then of course $$f_a(x_a)=0\implies a^{x_a}=x_a$$

In order for this to coincide with the unique minimum, we must have $$f_a^{'}(x_a)=0\implies \ln(a)x_a=1\implies x_a=\frac 1{\ln(a)}$$

Thus we are lead to solve $$a^{\frac 1{\ln(a)}}=\frac 1{\ln(a)}\implies a = e^{\frac 1e}\approx 1.44466786$$

Finally we see that $$a^x≥x\;\forall x \iff \boxed {a ≥ e^{\frac 1e}}$$

Why don't you write $$x\ln(a)\geq \ln(x)$$ or $$x\ln(a)-\ln(x)\geq 0$$ and now define $$f(x)=x\ln(a)-\ln(x)$$ if $$x>0$$

• For $x<0$ is $$a^x>x$$since $$a^x>0$$ for all real $x$ – Dr. Sonnhard Graubner Apr 9 '18 at 17:46
• Yea I deleted my comment sorry ^^ Writing things down I realized that you did the $ln$ transformation at an earlier step than him – Max Ft Apr 9 '18 at 17:48
• It is not a Problem,i thank your for your interesting discussion. – Dr. Sonnhard Graubner Apr 9 '18 at 17:49
• Oh ok. I also forgot that $a^x$ can only be positive if both $a$ and $x$ are reals. Thanks! – Max Ft Apr 9 '18 at 17:50

We can restrict ourselves to the case $a>1$ as the properties is false if $a \leq 1$.

• Then the case $f'$ has no zero never occurs as $x_0= -\frac{\ln(\ln(a))}{\ln(a)}$ has you have shown.
• For $x=x_0$ you have $a^{x_0}=\frac{1}{\ln(a)}$ so: $$f(x_0)=\frac{1}{\ln(a)}+\frac{\ln(\ln(a))}{\ln(a)}$$ so the sign of the minimum of $f$ is the sign of $\ln(\ln(a))+1$.