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
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
$y = a^y \implies y = e^{y \log(a)} \implies y e^{-y \log(a)} = 1$.
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 $\log_{10}(x)$.