# Proving the Dominance of e^x over x^e: A Mathematical Inquiry

• Echo 6 Sierra
In summary, our Cal-I Professor asked which is larger, e^x or x^e. It turns out that they are equivalent for all x except e. However, it is possible to find an interval where e^x is larger than x^e.
Echo 6 Sierra
This isn't HW and I don't know what form of math it it but it was posed to us by our Cal-I Professor. First, which is larger?

$$e^x$$ or $$x^e$$.

Yes, ^x. But he asks how we as individuals would prove this mathematically? All I can see is to plug in a number for x and go, but I don't think this is what he is after. Would I be hot or cold if I took the log of both and came up with xlne & elnx and then did something, anything else? Either way is just substitution and I'm just rambling now...a tall glass of Port and it's off to bed. :zzz:

These functions are equivalent for x = e, so did he define a certain interval for this proof? Not that it would matter in this analytical case, however it is important to know the full problem statement.

No, he didn't give an interval. He just asked how we would prove one would be greater than the other. I thought it might be something similar to how you would prove/grind out the law of sines, cosines, or a trig identity.

Thank you for the reply. E6S.

Last edited:
No, you can't "prove" one is strictly larger than the other because that's not true. As theelectricchild pointed out ex= xe if x= e. However, the graphs are tangent there. it's not too difficult to see that, for all x except e, xe< ex.

This reminds me of Knoebel's Exponentials Reiterated. I don't recall how Knoebell solves the general question $$x^y<y^x$$, but by graphing $$x^{1/x}$$ it's pretty clear that for all $$x\neq e$$ there are two points where $$x^y=y^x$$ for fixed x, with one larger inside and the other larger outside the open interval bounded by these two points.

There is a bit of a question of what your prof meant by larger. If he was just referring to the domain where both are real functions (ie the non-negative reals) then with some playing you get $$e^x = x^e \Leftrightarrow x^{1/x}= e^{(e^{-1})}$$. Then you can take the derivative of $$x^{1/x} = e^{ln(x)/x}$$. It's not too hard to show that this derivative is positive for x<e and negative for x>e. So you just need values a<e<b such that $$e^a > a^e$$ and same for b. Since both functions are continuous you are done. You could take a=1 and b=10 since $$e^{10} > 2^{10} > 10^3 > 10^e$$.

Hope that helps

Steven

Didn't make it to class last night to get this wrapped up. I'll email him and see if he can tell me how it came out in class.

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