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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?

Kamushek
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  • Taking the $8$th root of both sides and rewriting gives $e^{n\ln(2)/8}=n$ (restricted to the positive root due to the original equation) from which one can apply the above. For the second one, taking the square root gives $\pm10n=e^{n\ln(2)/2}$. – Simply Beautiful Art Sep 10 '19 at 20:21
  • If the desire is for numerical solutions, I would suggest fixed-point iteration, Newton's method, or bisection, as simple starting points. – Simply Beautiful Art Sep 10 '19 at 20:23
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    Yep, forgot that @SimplyBeautifulArt – Thomas Andrews Sep 10 '19 at 20:25
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    You have $n2^{-n/8}=1$ or $-\frac{n\ln 2}{8}e^{-n\ln 2/8}=-\frac{\ln 2}{8}$ or $-\frac{n\ln 2}{8}=-\frac{n\ln 2}{8}=W\left(-\frac{\ln 2}{8}\right)$ or $$n = -\frac{8}{\ln 2}W\left(-\frac{\ln 2}{8}\right)$$ where $W$ is the Lambert $W$-function. – Thomas Andrews Sep 10 '19 at 20:30
  • Thank you, all. I appreciate your help. But for now, I don't know what "Lambert W-function" is :D Guess it's to complex for me now :C – Kamushek Sep 10 '19 at 20:43
  • @ThomasAndrews, if u write it as an answer, I'll mark it :D (voted for all ur comments instead :D) – Kamushek Sep 10 '19 at 20:44
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    @Kamushek I can't, the question is closed. – Thomas Andrews Sep 10 '19 at 20:46
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    Google Lambert $W$ function. $W(z)$ is the solution of $we^w=z.$ – Thomas Andrews Sep 10 '19 at 20:47
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    In particular, if $-1/e<z<0$ there are two real values of $w.$ Since $-1/e<-\frac{\ln 2}{8}<0$ there are two values of $n.$ We see that there is a solution between $1$ and $2.$ and another solution between $32$ and $64.$ – Thomas Andrews Sep 10 '19 at 21:00

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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 $x=4$.

Alessandro Cigna
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