How to solve this logarithmic equation? $8n^2 = 64n\log n$, ($\log n$ here is base 2) I have tried to convert it to $n-8\log n = 0$, but how to solve the latest?
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Please use MathJax, see this link for a tutorial. – Paul Aljabar Jan 05 '18 at 12:59
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This doesn't have any solutions using elementary functions. But, using the Lambert W function, we get: $$n = -\frac {8}{\ln 2} \operatorname{W} \left (-\frac {\ln 2}{8} \right)$$ and $$n = -\frac {8}{\ln 2} \operatorname{W}_{-1} \left (-\frac {\ln 2}{8} \right)$$
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In case anybody has doubts concerning the usefulness of those expressions: Wolfram Alpha happily converts them into numbers: like here – Jan 05 '18 at 13:10
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Thanks for the reply :) sorry I don't know about Lambert function, will it allow me to get the final numeric value for n? e.g. n = 42 Just found some solutions to my exercise, but not sure which method is good to use http://atekihcan.github.io/CLRS/E01.02-02/ – Ani Alaverdyan Jan 05 '18 at 13:16
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If you cannot use Lambert function, consider that you look for the zero's of $$f(x)=x-8\log_2(x)$$ The first derivative $$f'(x)=1-\frac{8}{x \log (2)}$$ cancels for $x_*=\frac{8}{\log (2)}$ and $$f(x_*)=\frac{8-8 \log \left(\frac{8}{\log (2)}\right)}{\log (2)}\approx -16.6886$$ The second derivative test shows that this corresponds to a minimum. So, there are two roots to the equation.
If you plot the function, you will see that the roots are close to $1$ and $40$. So, start Newton method and below are given the iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.000000000 \\ 1 & 1.094862617 \\ 2 & 1.099983771 \\ 3 & 1.099997030 \end{array} \right)$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 40.00000000 \\ 1 & 43.61991000 \\ 2 & 43.55927562 \\ 3 & 43.55926044 \end{array} \right)$$
Claude Leibovici
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