how to solve the logarithmic equation which has both n and logn 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?
 A: 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 $$n = -\frac {8}{\ln 2} \operatorname{W}_{-1} \left (-\frac {\ln 2}{8} \right)$$
A: 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
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1.000000000 \\
 1 & 1.094862617 \\
 2 & 1.099983771 \\
 3 & 1.099997030
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 40.00000000 \\
 1 & 43.61991000 \\
 2 & 43.55927562 \\
 3 & 43.55926044
\end{array}
\right)$$
