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The given equation: $$\frac{8} {\ln(2)} \cdot \ln(x) = x.$$

Some algebra, and then we get the solutions:

$$x = e^{W_0(-\frac{\ln(2)} {8})}\approx 1.1 ,\;\; x = e^{W_{-1}(-\frac{\ln(2)} {8})} \approx 43.5593.$$

I found $W_{0}$ with Taylor series.

But there is addition real solution, $W_{-1}$.

How I find the $W_{-1}$?

Gary
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NateD
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  • You could just edit your original question instead of posting a new one. You may use http://dlmf.nist.gov/4.13.E6 for the other branch with $$ t = - \sqrt {2\left( { - 1 - \log \left( {\frac{{\log 2}}{8}} \right)} \right)} . $$ – Gary Oct 24 '21 at 11:12
  • Btw, how is $W_0 = e^W$? – Gary Oct 24 '21 at 11:17
  • Sorry, i meant $x = e^{W}$ The $W_{0}$ is to say that for this x i apply $W_{0}$ and not the $W_{-1}$ – NateD Oct 24 '21 at 11:21
  • Thanks @Gary . but I'm not fully understand how did I find the $C_{n}$ ? and how did you get to this this t?. is there any simpler series representation for $W_{-1}$ with n's and constants, like we found the $W_{0}$ ? – NateD Oct 24 '21 at 12:19
  • If you look below the formula I linked, there is a recurrence for the coefficients. There is no simple series for $W_{-1}(z)$ like the one for $W_0(z)$ because it is not analytic at $z=0$. With my $t$, $$
    • e^{ - 1 - t^2 /2} = - \frac{{\log 2}}{8}

    $$ and that is what you wanted.

     – Gary Oct 24 '21 at 12:26

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