How to solve this logarithmic equation for $x$? 
Given the following equation $$20 x \, \log_2(x) = 10^9$$ How to solve for $x$?

I have tried to solve it, but I get something like
$$20^{50000000 } = n^n$$
and I'm stuck there. Could you please give me step-by-step instructions? Thank you.
 A: Welcome to the world of Lambert function !
The solution of $n^n=k$ is given by $$n=\frac{\log (k)}{W(\log (k))}$$ The Wikipedia page gives approximations for large values of the argument
$$W(x)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(-2+L_2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.
If you apply it using $x=50000000 \log (2)$, you will get something quite close to the soution which is $\approx 2.36\times 10^6$.
If you cannot use Lambert function, only numerical methods will do the job.
Considering the function $$f(x)=\frac{20 x \log (x)}{\log (2)}-1000000000$$
$$f'(x)=\frac{20 \log (x)}{\log (2)}+\frac{20}{\log (2)}$$ let me be very lazy starting Newton iterations using $x_0=1000$. The generated iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1000 \\
 1 & 4.382831512\times 10^6 \\
 2 & 2.396102501\times 10^6 \\
 3 & 2.361694528\times 10^6 \\
 4 & 2.361678691\times 10^6 
\end{array}
\right)$$
A: Well,\begin{align}
20x \log_2 x  &= 10^9 \\
\implies \log_2 x^x &= 5 \times10^7\\
\implies x^x &= 2^{5 \times 10^7} \\
\implies x^x &= 2^{50000000}
\end{align}
The solution of which, as @Arthur notes, will be in the form of the Lambert $W$ function.
A: $$x\log x=\frac{\log2\cdot10^9}{20}$$
immediately gives
$$\log x=W\left(\log\sqrt2\cdot10^8\right).$$
By Alpha:
$$x=2.361678691102962518047152935761346465304830445206392\cdots × 10^6$$
