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I want to find a way to solve $x^9=3^x$ analytically, for two roots. one of them can be found below $$x^9=3^x\\(x^9)^{\dfrac {1}{9x}}=(3^x)^{\dfrac {1}{9x}}\\x^ { \ \frac 1x}=3^{ \ \frac 19}\\x^ { \ \frac 1x}=(3)^{ \ \frac 3{27}}\\x^ { \ \frac 1x}=(3^3)^{ \ \frac 1{27}}\\ x^ { \ \frac 1x}=(27)^{ \ \frac 1{27}}\\\to x=27$$ but there is a second root if we take a log of both sides and graph them $$x^9=3^x \to \log x^9 =\log 3^x \to 9\log x=x\log 3$$ enter image description here

With respect to the picture, there must be $x=1.151$ as another solution, but how can we find it? (without using numerical or graphical methods)
I do appreciate any hint. Thanks

Khosrotash
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    You'll need the Lambert W function (cf. https://en.wikipedia.org/wiki/Lambert_W_function). – Abezhiko Feb 21 '24 at 19:39
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    You may use these answers and different branches if needed – Тyma Gaidash Feb 21 '24 at 19:40
  • @ТymaGaidash و@Abezhiko : thank you both. it helps me. – Khosrotash Feb 21 '24 at 20:08
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    But let me point out that while the Lambert $W$ function provides you a way of writing out an expression for the value in the form of $x = ...$, to actually evaluate that expression requires using numerical methods. – Paul Sinclair Feb 23 '24 at 15:54
  • @PaulSinclair: Yes, I agree with you. So the result is : There is no analytic solution ? – Khosrotash Feb 23 '24 at 16:10
  • More that in general, there is no such thing as an "analytic solution". Once you get past a finite combination of $+,-,\times,\div$, everything else must be calculated by numerical methods. All that using such things as $\sqrt x, \sin x, \cos x, e^x,\ln x, W(x)$ does is to hide the numerical methods used to evaluate them under the carpet. To be sure, we have some nice algorithms that help in evaluating them, and have been built into applications so you don't have to do the methods yourself. But they are still numerical methods. – Paul Sinclair Feb 23 '24 at 16:19
  • For this example, using Newton's method to find $x$ straight off would give you as accurate an answer probably more easily than converting it to use $W$, and finding some algorithm for $W$ and calculating that algorithm. Now if you are wanting to do analysis on the solution to see how it behaves as you change the inputs - for that $W$ is what you need. But actually calculating values? A simple numerical method such as Newton's is often the best approach. – Paul Sinclair Feb 23 '24 at 16:21
  • Lambert W has expressions in terms of sums and products here, which give analytic/exact forms, depending on what you classify them to be. – Тyma Gaidash Feb 23 '24 at 16:57
  • @ТymaGaidash - You cannot actually perform an infinite product or sum. Anything expressed as such can only be approximated - i.e. calculated by "numerical methods". – Paul Sinclair Feb 23 '24 at 20:57
  • @PaulSinclair There was a small typo; there is also an integral representation – Тyma Gaidash Feb 23 '24 at 23:28

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