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Solve for $x$: $(x^x)^{2015}=2015$

Tried several times, but have no idea about how to start even.

Harry Peter
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dsh
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  • Might I ask, is this a question from a contest? If so, which one? – muaddib Jun 09 '15 at 15:09
  • @muaddib to be honest it is not a contest, it is a high school exam – dsh Jun 09 '15 at 15:10
  • Try $x^x=2015^\frac 1{2015}$ and solve numerically. – Hypergeometricx Jun 09 '15 at 15:14
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    @kareem have you heard of the Lambert W function? I'd be shocked if you encountered this function at the high school level, but it could help you solve the equation. http://en.wikipedia.org/wiki/Lambert_W_function#Example_2 – graydad Jun 09 '15 at 15:15
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    You get $2015^{1/2015} \approx 1.00378300579661$ and so slightly less $x \approx 1.00376877529684$ – Henry Jun 09 '15 at 15:15
  • @Henry how did you get x=1.....? Thanks – dsh Jun 09 '15 at 15:28
  • I started with a good but not exact approximation is $2015^{\left(2015^{-2016/2015}\right)} \approx 1.00376872176686$ and then adjusted – Henry Jun 09 '15 at 15:29
  • Oh, but is it possible to find some proper solution, without approximation – dsh Jun 09 '15 at 15:33
  • I think that you have misread the statement: it is impossible to be required to compute the solution of such an equation at high-school level. In fact, I think that you are only required to prove that there exist a unique solution, not to also compute it. – Alex M. Jun 09 '15 at 15:35
  • It is okay to compute it using any level, because I sort of need to prove to teacher, that its not school level – dsh Jun 09 '15 at 15:38
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    $\big(x^x\big)^{\color{red}-2015}=2015$ admits $x=\dfrac1{2015}$ as solution. – Lucian Jun 09 '15 at 18:11
  • @Lucian Nice observation. It seems to suggest a method of attack for solving Op's problem, but maybe not, if integration is a good analogy changing terms by even a sign can yield un-evaluable expressions. – Zach466920 Jun 09 '15 at 19:58

2 Answers2

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Here's a start $$(1) \quad (x^x)^{2015}=2015$$ $$(2) \quad x^{2015 \cdot x}=2015$$ take the natural log of (2)... $$(3) \quad (2015 \cdot x) \cdot \ln(x)=\ln(2015)$$ $$(4) \quad \ln(x)={{\ln(2015)} \over {2015 \cdot x}}$$ finally, $$(5) \quad x=2015^{(2015 \cdot x)^{-1}}$$ What is this you might be wondering? Using the theory of dynamical systems we can see that the derivative of this function on the right side of (5) is less than 1. This means iterating (5) will approach the solution of (1)!

Thus the solution is...

$$x=2015^{\left(2015 \cdot 2015^{\left(2015 \cdot 2015^{\left(2015 \cdot 2015^{\left(2015....\right)^{-1}} \right)^{-1}} \right)^{-1}} \right)^{-1}}$$

This converges extremely fast. Just one convergent, $2015^{{1} \over {2015}}$, yields $x=1.00378...$. Very close indeed!

Zach466920
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$$x^x=\sqrt[2015]{2015}\to x=e^{W(\ln(2015)/2015)}.$$