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$5^x = x^5$

I'm having trouble solving this problem. I tried taking log of both sides to get: $$x\log(5) = 5\log(x)$$ then $$x/\log(x)=5/\log(5)$$ But I'm not sure where to go from here. Am I in a wrong direction?

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  • $\begingroup$ @Dovah-king, Is that the only solution? $\endgroup$ – math Mar 3 '17 at 1:23
  • $\begingroup$ @Dovah-king not for $x$ real; there is a solution between $1$ and $2$. $\endgroup$ – Joffan Mar 3 '17 at 1:24
  • $\begingroup$ Yeah, i was thinking about a solution between $1$ and $2$. Is there an analytic method for solving this? $\endgroup$ – embedded_dev Mar 3 '17 at 1:24
  • $\begingroup$ @Joffan, How do you know that there is a solution between 1 and 2? $\endgroup$ – math Mar 3 '17 at 1:25
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    $\begingroup$ @math why do we know there is a solution between $1$ and $2$? because both $5^x$ and $x^5$ are continuous for positive $x$. At $x=1$ you have $5^1>1^5$ meanwhile at $x=2$ you have $5^2<2^5$ (as $25<32$) so they must have crossed (at least once) on that interval as per the intermediate value theorem. $\endgroup$ – JMoravitz Mar 3 '17 at 1:27
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The other real solution is $$-\frac{5}{\ln 5} W\left(-\frac{\ln 5}{5}\right)$$ where $W$ is the Lambert W function.

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  • $\begingroup$ And Wolfy says this is about 1.76492. $\endgroup$ – marty cohen Mar 3 '17 at 2:16
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Hint:

From the variations of the function $\dfrac{\ln x}x$, you can show there are two solutions: $x=5$, of course, and a solution between $1$ and $\mathrm e$.

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