I've tried a few different approaches but I'm not getting anywhere with this.
100log(n) = n^(0.5) ==> log(n) = n/100 ==> n = 2^(n/100) Stuck at this dead end.
100log(n) = n^(0.5) ==> log(n^(100)) = n ==> n^(100) = 2^(n) Stuck again.
I also looked at this post: $4\log_2(n)=n$ How to Solve for $n$?. However, none of the answers there helped me out much. The first answer suggests using, "the Lambert W-function, which can't be solved with elementary functions (but can be expressed as an infinite sum using the Lagrange inversion forumula)." But then it continues to say, "However just by trying random powers of 2 you can get an integer solution of n=16, probably the easiest way." Trying random powers of 2 won't work in this case unless I have a lot of time to spare and I've never heard of the Lambert W-function or Lagrange inversion formula. Has anybody had success with the Lambert W-function or Lagrange inversion formula in this case? Open to any other suggestions as well.
In the same post I tried playing around with the 3rd answer from the top but that didn't help me out either.
Any help with this would be greatly appreciated.