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$$\frac{x^x + x^x + x^x }{x + x + x} = 9$$ $$x^{-x} = ?$$

This problem seems very complex to me. I've found the answer as $\frac {1}{27}$. However, I don't think that I've found the correct answer. Can you assist?

Goendo
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    Please share your workings, within your post. It means very little to say "I found xyz to be the answer, but I think it's wrong". Show us how you found what you found. – amWhy Feb 26 '18 at 21:35
  • Please tell us how you arrived at $\frac{1}{27}$, otherwise your question will probably be closed for lack of context, and that way we can also tell you where you've gone wrong (if you have). – Henrik supports the community Feb 26 '18 at 21:35
  • @amWhy Okay, I will. My attempt: $$\frac {3.x^2}{3.x} = 9$$ Here we get $$x^{-x} = \frac {1}{27}$$ – Goendo Feb 26 '18 at 21:36
  • Where does the $3^\cdot x^2$ in the numerator come from? Note $x^x+ x^x+x^x = 3x^x$. But perhaps you simply have a typo in your comment. Now, how did you determine your answer of $\frac 1{27}$?. – amWhy Feb 26 '18 at 21:41
  • @amWhy That was just chance to get that value, now I know how to determine that. – Goendo Feb 26 '18 at 21:42

3 Answers3

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Hint:

$$9 = \frac{x^x+x^x+x^x}{x+x+x} = \frac{3x^x}{3x} = \frac{x^x}{x} = 9.$$

I don't know of a way to solve this analytically, but a guess of $x=3$ shouldn't be too hard to make from here.

Note: There is actually another solution at $x\approx 0.0895224$ which gives $x^{-x}\approx 1.24115$. Because of this, it is not likely than an analytical solution is possible, nor is there a unique answer to the question. It is likely that the question writer(s) intended this to be a "guess-and-check" sort of problem.

Carl Schildkraut
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Note that

$$\frac{x^x + x^x + x^x }{x + x + x} =\frac{3x^x}{3x}=x^{x-1} = 9\implies x=3$$

to justify the solution and check for others we can consider

$$x^{x-1} = 9\iff \log x^{x-1} = 2\log 3\iff (x-1)\log x=2\log 3$$

and study the zeros for the function

$$f(x)=(x-1)\log x-2\log 3$$

Notably note that

$$f'(x) = -\frac1x +\log x +1=0$$

and to study this we can consider

$$g(x)=f'(x) \implies g'(x)=\frac{x+1}{x^2}>0 \quad \forall x>0$$

then f'(x) may have only one zero that is $x=1$ by inspection.

It easy to show, with a few of work by derivatives, EVT and limits that this is a negative minimum for $f(x)$ and that another (and only one) solution have to exist in the interval $x\in(0,1)$.

user
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  • Except that 3 is not the only root. – Did Mar 04 '18 at 15:33
  • @Did did you read the full answer? – user Mar 04 '18 at 15:35
  • Of course, why do you ask? Are you going to cling to your post at all cost, although it does not answer the question? FYI, the implication sign in the first displayed line is w.r.o.n.g. – Did Mar 04 '18 at 15:38
  • @Did Don't read as an implication but as "we find that x=3" and then read "to justify the solution and check for others..." I'm confident that you can understand my aswer if you read it better! Bye – user Mar 04 '18 at 15:41
  • Ah. You use "$\implies$" on a math site but this symbol should not be read "as an implication"? Are you serious? – Did Mar 04 '18 at 15:42
  • @Did If you read also the following line the meaning of the symbol used is clear, you always need to consider the context when you read something, open your mind! – user Mar 04 '18 at 15:47
  • My mind is quite open to every kind of mathematics.... except the wrong ones. Sorry about that (or not). Unsurprisingly, we now know the answer to my former question: "Are you going to cling to your post at all cost?" – Did Mar 04 '18 at 15:51
  • @Did If you don't agree with any of my answer you can let a comment or give a suggestion or simply downvote, MSE works in this way if I'm not wrong. Please avoid to bothering me with useless and stupid considerations. Be quite and spend your time for something else. Thanks again for your opinion. Bye – user Mar 04 '18 at 15:55
  • This is exactly what I did with my first comment, hmmm? – Did Mar 04 '18 at 15:57
  • @Did Your way of giving advice is not very polite and very annoying to me, as your way to write comments in bold style is very silly, you should learn also a bit of education as well as some theorems you know. – user Mar 04 '18 at 16:07
  • Are you sure you are in the position of giving etiquette lessons to others after your "performance" in this comment thread? "Bye". – Did Mar 04 '18 at 16:42
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We have $$x^x=9x$$

Since, $f(x)=x^x$ is a convex function, we see that this equation has two roots maximum.

But, $3$ is a root and there is a root $x_1$, where $0<x_1<1$.

For $x=3$ we obtain $x^{-x}=\frac{1}{27}.$

Also, $$x_1^{-x_1}=1.24...$$

Michael Rozenberg
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