Solve equation like $f(x) = \log_2(g(x))$

Question

I've tried to solve two equations, and they look similar to me. $$f(x) = \log_2(g(x))$$ In the original task, I should compare them, but I wanna solve them as equations. Is it possible using analytical method? $$f(x) = \log_2(g(x))$$ Equations: $8n^2 = 64n\log_2(n)$ and $100n^2 = 2^n$
I'll give my thoughts with the first one. $$8n^2 = 64n\log_2(n)$$ $$n = 8\log_2(n)$$ $$\log_2(2^n) = \log_2(n^8)$$ $$2^n = n^8$$ And now I'm stuck, does anybody have some suggestions?

Answer 1

There isn’t an analytical method but you could find particular integer solutions and study the associated real valued function to know number of solutions.

For example if you’re looking for solutions of $n^8=2^n$, you can see that $n^8$ and so $n$ must be a power of $2$. So suppose $n=2^a$, we want $2^{8a}=2^{2^a}\iff 8a=2^a\iff 2^{a-3}=a$ but by induction you can see that for all $a\ge 6$ then $2^{a-3}>a$ and for $a<6$ by inspection you see there aren’t solution.

If you wanted the solutions of, for instance $2^n=n^2$ you could see that $n=2$ and $n=4$ are solutions, to show they are unique you must recur to calculus: calling $f(x)=2^x-x^2$ we see that $f’(x)=2^x\ln 2-2x$ and $f’’(x)=2^x\cdot \ln^2 2-2>0$ for $x\ge 3$, so in $[3,\infty) $ we could have at most one solution, that is $x=4$.