# Solving logarithmic equation for $n$

I have the following equation and I am trying to isolate $n$:

$$8n^2 = 64 n\log_2 n$$

Haven't done algebra in years and can't figure out how to get rid of the $\log_2$.

• If you write it as $8\times n\times n=8\times n\times 8\log_2n$, I hope you can see that if $n\neq0$, your equation is equivalent with $n=8\log_2n$ which should be easier to manipulate. You have to check the case $n=0$ separately. – Joonas Ilmavirta Jan 8 '15 at 10:01
• Yea I can get to n=8log2n but thats as far as I make it. – moesef Jan 8 '15 at 10:02
• There is no elementary way of solving equations that contain the unknown and a logarithm of the unknown. Assuming $n>0$, you have $n=8\log_2n\iff n=\log_2n^8\iff 2^n=n^8$. The only solution is $\approx1.1$ (closed form expressions involve a special function). (Tip: When asking in the future, tell what you already know yourself. This saves others from explaining something you already know.) – Joonas Ilmavirta Jan 8 '15 at 10:16
• ...and there is also a second solution, near 44. But these two are all. You can also try using WA. – Joonas Ilmavirta Jan 8 '15 at 10:25
• Sure. Thanks for the answer. If you put that exact comment in an answer I'll mark as closed. – moesef Jan 8 '15 at 10:25

As said in comments, because of the logarithm, $n \gt 0$; so, as you wrote, the equation simplifies to $$n = 8 n\log_2 n$$ which does not have a solution in terms of elementary functions.
However, any equation which can be written as $$a+b x+c \log(d+ex)=0$$ has a solution which is expressed in terms of the Lambert function. In your case, there are two solutions given by $$n_1=-\frac{8 }{\log (2)}W\left(-\frac{\log (2)}{8}\right)\approx 1.099997030$$ $$n_2=-\frac{8 }{\log (2)}W_{-1}\left(-\frac{\log (2)}{8}\right)\approx 43.55926044$$
If you do not want (or cannot) to use Lambert function, you could use a root-finding method such as Newton. Starting with a "reasonable" guess $n_0$, this will find the solution of $$f(n)=n - 8 n\log_2 n=0$$ updating the guess according to $$n_{k+1}=n_k-\frac{f(n_k)}{f'(n_k)}$$ For your case, the iterative schme will then be $$n_{k+1}=\frac{8 n_k (\log (n_k)-1)}{n_k \log (2)-8}$$ Let us start with a very poor estimate such as $n_0=20$; Newton successive iterates will then be $54.4636$, $43.8990$, $43.5597$, $43.5593$ which is the solution for six significant figures.