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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.

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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 $$\left( \begin{array}{cc} n & x_n \\ 0 & 1000 \\ 1 & 4.382831512\times 10^6 \\ 2 & 2.396102501\times 10^6 \\ 3 & 2.361694528\times 10^6 \\ 4 & 2.361678691\times 10^6 \end{array} \right)$$

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  • $\begingroup$ Hi, Claude. If $x\log x=y$ then $y\approx x/\log x$. In this case, this gives the starting approximation $y=2\cdot10^6$. $\endgroup$ – Yves Daoust Oct 9 '17 at 7:36
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    $\begingroup$ @YvesDaoust. Hi Yves ! I know but, as I wrote, I wanted to be lazy to show how well would work Newton method. Cheers. $\endgroup$ – Claude Leibovici Oct 9 '17 at 7:46
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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.

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$$x\log x=\frac{\log2\cdot10^9}{20}$$

immediately gives

$$\log x=W\left(\log\sqrt2\cdot10^8\right).$$

By Alpha:

$$x=2.361678691102962518047152935761346465304830445206392\cdots × 10^6$$

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