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As the title says,

We seek the smallest real number $a\gt 0$ for which the equation $a^x=x$ has no real solutions.

This is inspired by this question.

I must admit that I did not have much luck with this..Any suggestions?

3 Answers3

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This is my solution, which I can't guarantee is correct or as rigorous as one would like.

First note that for $a \le 1$ it is trivial to see that real solutions exist, so we can focus on the $a > 1$ case.

The idea then is to determine when the curves $a^x$ and $x$ intersects, which is exactly when the equation has solutions. So first reformulate it as a function and calculate it's derivatives:

$$ f(x) = a^x - x \\ f'(x) = ln(a)a^x - 1\\ f''(x) = ln(a)^2a^x $$

Then use the derivatives to calculate $x_{min}$ for $f$, which is just number crunching and you get: $$ -\frac{\ln(\ln a)}{\ln a} $$ Then insert it into $f$ to get its $y_{min}$: $$ \frac{1 + \ln(\ln a)}{\ln a} $$ So if $y_{min} > 0$ the curves does not interest and no real solutions exist.

$$ \frac{1 + \ln(\ln a)}{\ln a} = 0 \\ \iff \ln(\ln a) = -1 \iff a = e^\frac{1}{e} $$

In other words, solutions exist iff $0 < a \leq e^\frac{1}{e}$.

2

Obviously, $x\ge0$. From the equation, we draw

$$a=\sqrt[x]x.$$

This function has a single maximum at $x=e$, so that

$$a\le \sqrt[e]e.$$


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As pointed out in the comments, this solution is not quite accurate. See Björn Lindqvist's solution.



Let $f(x) = a^x$. The idea is that we need to find the value of $x$ such that $f'(x) = 1$ and this point $x$ equals $a^x.$ The derivative is $f'(x) = \ln(a)a^x$. Setting this equal to one and solving for $x$ we see that

$$ x = -\ln(\ln(a))\;. $$

So our point is $(-\ln(\ln(a)),a^{-\ln(\ln(a))})$, and we need to lie on the line $y=x$, so we need the solution to the equation

$$ -\ln(\ln(a))\;\;=\;\;a^{-\ln(\ln(a))}\;. $$

I'm not sure if there is a way to solve for $a$ explicitly, but putting this into WolframAlpha we get the approximation $a \approx 1.27627610348955$. WolframAlpha, OEIS, and a brief Google search indicate that this isn't some known (or popular) constant.

Mike Pierce
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    Thank you for this! I was actually hoping that this was indeed a "known" constant of some worth. If there is a way to solve explicitly, it would be nice to have more info on it. Thanks again.. – MathematicianByMistake Oct 26 '16 at 21:30
  • But if $a$ is 1, $1^x = x$ has the solution 1. Doesn't that contradict your answer? – Stand with Gaza Mar 05 '17 at 03:21
  • @BjörnLindqvist I don't think so. That constant in my answer is the smallest real number $a$ such that $a^x=x$ has no real solutions. For any value of $a$ less than that constant, it's fine for that equation to have solutions. – Mike Pierce Mar 05 '17 at 03:40
  • You are right. But now I think there are some a's that are larger than what you have calculated that have real solutions. E.g $1.4^x = x$ is solvable. I think $a$ should be $e^{1/e}$. – Stand with Gaza Mar 05 '17 at 12:01
  • @BjörnLindqvist yeah, just looking at the graphs of the curve $e^{x/e}$ and $x$ you are definitely correct. I'll try to figure out exactly what's wrong with my approach and calculation later, but until then you should type up what you've got as an answer. – Mike Pierce Mar 05 '17 at 14:48