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I am curious how to solve equations like $15^x = x^{16}$ or $x^x = 1991$. Yes, of course Wolfram Alpha or Matlab can calculate it, or it can be approximated (as we learned in school). But are there any other solution methods? They can be transformed to a form where on one side there is $x\ln(x)$ or $x/\ln(x)$. How to solve them? What if we have $15^x+16^x+x^x-x^{17}-x^{18} = 0$? ($x>0$ in all equations).

Zev Chonoles
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lajos.cseppento
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    This should be helpful. – L. F. Sep 01 '13 at 13:43
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    You could use numerical methods like the Newton-Raphson method. – Mufasa Sep 01 '13 at 13:44
  • @Mufasa That numerical method is fine to find the solutions, but the workout would be very cumbersome! I believe L.F.'s suggestion is the best method to go for. – NasuSama Sep 01 '13 at 13:57
  • What does solution means in this context? Galoise anyone? – jimjim Sep 01 '13 at 14:52
  • @DanielR : First thing is to be clear about the question, "one unknown equation" is misleading, as all the "one unknown equation"s are easily solvable ( because the meaning of "one unknown equation" is "one unknown linear equation"). Once that we are clear what we are talking about then we can talk about why this "one unknown equation" can not be solved. The tag is now changed from (problem-solving) to transcendental equations. – jimjim Sep 01 '13 at 14:28
  • First, I wasn't aware that the term "one unknown equation" was defined to be a linear equation with one unknown. Why wouldn't $x=\sin x$ be an equation with one unknown? It is! Can you point me to a reference on the subject?

    Second, I'm pretty sure that this kind of answer doesn't really help the asker. It's pretty obvious what kind of equations he's referring to. Why not put it as a comment instead?

    Third, are you from Värmland by any chance? :)

     – Daniel R Sep 01 '13 at 14:37
  • @DanielR : googling "one unknown equation" gives the first 5 results including "linear" in their titles, or their context. $ x = \sin (x)$ is not a one unknown equation because $\sin (x)$ has infinitely many terms of powers of x in it, and an equation with infinite many terms is not usually considered solvable, This kind of answer clears up the misunderstanding of what one unknown equation is, and what a solutions means ( Galoise anyone?) and finaly I am from Canberra! :) Thanks Daniel – jimjim Sep 01 '13 at 14:49
  • @Arjang: Your answer seems to have been turned into a comment. Anyway, I think you are the one who are mixing up the terminology here. Galois theory gives that there are polynomial equations (with degree $\geq 5$) whose solutions cannot be expressed by radicals. That doesn't mean that there are no solutions to a those equations. And, since you refer to Galois, what about a quartic equation, which does have solutions expressable in radicals? Is this a linear equation with one unknown in your book then? And what about $\sin x = \cfrac12$? – Daniel R Sep 01 '13 at 17:00

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One option is to find the solution in terms of Lambert $W$ functions. Lambert $W$ function is defined as: $$ W(x)e^{W(x)}=x. $$ Now for $15^x=x^{16}$, you can make an equivalent equation which is: $$ 15^{\frac{-x}{16}}x=1 $$ $$ e^{x\frac{-\ln15}{16}}\left(x\frac{-\ln15}{16}\right)=\frac{-\ln15}{16} $$ And then you get: $$ x=\frac{-16}{\ln15}W(\frac{-\ln15}{16}). $$ For $x^x=1991$, as you said it yourself, you can turn it into:

$$ e^{\ln x}\ln x=\ln(1991)\rightarrow x=e^{W\left(\ln1991\right)}. $$ Now for the case of sum of polynomials and mixture of exponential function, the current form of Lambert $W$ function will not work. Look at the generalization of Lambert function in Wikipedia.

Arash
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