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I'm trying to calculate the range of $a$ that satisfies $a^x>x$ for any real number $x$. Reduce[a^x > x, a] fails. If it's not possible to get a symbolic answer, a numerical solution is also okay.

An answer showed Reduce[a^x > x, a, Reals], and it does help, but I need a constant range of $a$ not relative to $x$. For example, $2^x>x$ is true for any real $x$, but $1.1^x>x$ is false. How can I calculate the constant point from which the inequality holds true?

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  • $\begingroup$ The obvious Resolve[ForAll[x, a^x > x], Reals] is unable to resolve this question. $\endgroup$
    – Roman
    Nov 12, 2022 at 9:37

3 Answers 3

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Observe the result of Reduce as given by user64494:

Reduce[a^x > x, a, Reals]

(x == 0 && a < 0) || (x <= 0 && a > 0) || (x > 0 && a > x^(1/x))

Note that the critical question is "what is the minimum possible value of $a$ such that $x^{1/x}$ is always less than $a$?" This minimum will occur at some variety of critical point of $a=x^{1/x}$.

Solve for x:

Solve[a == x^(1/x), x]

{{ x -> -ProductLog[-Log[a]]/Log[a] }}

Plug in a few values and notice that for small values these x values are strictly real (meaning there is a crossing point in the reals where this equality is satisfied and as such $a > x^{1/x}$ is not satisfied), and large values they are complex (meaning the equation is never satisfied with a real $x$ value).

By intuition, I assumed that the critical point is Exp[1/E]. I will leave it to someone more familiar with the intricacies of ProductLog to explain why.

Observe Plot[{Exp[1/E]^x, x}, {x, -5, 5}] and note that it contacts at exactly one point. For larger $a$ values than Exp[1/E], it will clearly always be greater than $x$, and for smaller values it will cross more than once.

plot of power of critical a versus x

Edit:

Naturally, re-reading one's own answer can lead to dramatically simpler insights. I say earlier that "what is the minimum possible value of $a$ such that $x^{1/x}$ is always less than $a$". Naturally, this is the maximum of $x^{1/x}$:

Maximize[x^(1/x),x]

E^(1/E)

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By default a and x are assumed complex. The Reals domain helps

Reduce[a^x > x, a, Reals]

(x == 0 && a < 0) || (x <= 0 && a > 0) || (x > 0 && a > x^(1/x))

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  • $\begingroup$ Thanks, but I need something slightly different. Please have a look at the edit. $\endgroup$
    – xiver77
    Nov 11, 2022 at 22:50
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As noted in the comment, the general

Resolve[ForAll[x, a^x > x], Reals]

Does not work. For specific values of $a$ we can define

f[a_?NumericQ] := Boole@Resolve[ForAll[x, a^x > x]]

and then plot

Plot[f[a], {a, 1, 2}, GridLines -> {{E^(1/E)}, {}}]

enter image description here

which gives an idea of the range of values of $a$ for which the statement is true, and confirms @eyorble's answer.

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