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A value of $y=.5295431$ does satisfy the equation

$$ y = (\log 2)^{y}$$

But I havn't seen any ways to prove it.

$\log$ is base $10$ and $\ln$ is $\log$ to the base $e$

Note: I would like to see a proof

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Do you mean you want to show that 0.5295431... is a solution to $y=(\log 2)^y$, or that you want to show that a solution to $y=(\log 2)^y$ is a solution to the equation in the title? – Matthew Conroy Mar 14 '12 at 0:06
When you say $log2$ you seem to mean $\log_{10}(2)$ – Henry Mar 14 '12 at 0:06
@Henry Yes that is correct – Kirthi Raman Mar 14 '12 at 0:14
@Matthew $(0.30103)^{(0.529543)} = 0.529543$ checked on Wolfram Alpha. – Kirthi Raman Mar 14 '12 at 0:15
That doesn't make clearer what you want to prove. You say "prove it", but it is not clear what "it" is: do you want help finding a numerical solution to $y=(\log 2)^y$ or do you want help showing that a solution to $y=(\log 2)^y$ is a solution to your equation in the title? – Matthew Conroy Mar 14 '12 at 0:21
up vote 4 down vote accepted

$y = a^y \implies y = e^{y \log(a)} \implies y e^{-y \log(a)} = 1$.

This gives us $$-y \log(a) e^{-y \log(a)} = -\log(a).$$ Setting $z = -y \log(a)$, we have $z e^{z} = -\log(a)$.

The solution to $ze^z = x$ is given by the Lambert W function, $W(x)$.

In your case, $x = -\log(a)$ and hence $z = W(- \log(a))$. Since $y = - \frac{z}{\log(a)}$, we get $$y = -\frac{W(-\log(a))}{\log(a)}$$

In your case, $a = \log_{10}(2)$. This gives us $y \approx 0.529543166 \ldots$.

EDIT: $\log(x)$ typically denotes $\log_e(x)$. Explicitly specify, if it is $\log_{10}(x)$.

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Something is not right $(0.30103)^{(0.757558)} = 0.40273 \neq 0.757558$ – Kirthi Raman Mar 14 '12 at 0:22
You might be taking to the base $e$? – Kirthi Raman Mar 14 '12 at 0:25
I think if I take to base 10 in your last expression I am getting the right answer.Sivaram, please do one favor. Show more steps, and how you got Lamber W function. Can you elaborate please. – Kirthi Raman Mar 14 '12 at 0:29

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