How to solve equations like $8n^{2} = 64n\log_2(n)$

Question

How do I solve this equation: $$ 8n^{2} = 64n\log_2(n) $$ I've plotted logarithmic curves for both functions and found that they intersect at $n = 44$, but I've no idea how to solve the equation.

Answer 1

You have a polynomial equation in $n$ and $log_{2}(n)$. For all $n\neq 1$, the both are algebraic independent. That means, $n$ is transcendental. Use the Lambert W function. Maple gives $n=-\frac{8LambertW(-\frac{1}{8}ln(2))}{ln(2)},-\frac{8LambertW(-1,-\frac{1}{8}ln(2))}{ln(2)}$. Type in Wolfram Alpha: Solve(8*n^2=64*n*log2(n),n).

Answer 2

$8n^2=64n log_2(n)\iff\frac n8=$ $log_2 (n)\iff 2^{\frac n8}=n$

So we have $(\sqrt[8] 2)^n=n$. This is a trascendental equation which can be solved by approximations or by computer (which is approximated too). I rather try the approximation by hand in this particular case.

$\alpha=\sqrt[8] 2=1.090507733$ so one has $\alpha^n=n$ to solve.

It is clear that $n$ is close to $1$ but can be also great in order to get the equality. I calculate only the solution near to $1$

We have $\alpha^1=1.09…..\gt 1$

$\alpha^{1.1}=1.09997314….\lt 1.1$

$\alpha^{1.009}=1.09....\gt 1.009$

$\alpha^{1.11}=1.10095...\lt 1.11$

We take the approximation $n=1.1$