2

I have two equations. I wasn't sure how to properly proceed with them.

  1. $ x(\log_8x - 1) = 64 $

  2. $ u(u-1) = \ln(u)$

I don't think I can set each argument to the right side like:

$u = \ln(u)$

$u-1 = \ln(u)$

The top equation is easy to approximate $x = 64$, but I'm looking for a more rigorous way. (Edit: I made a mistake on (1) it is now written correctly.

N. F. Taussig
  • 76,571
Frederick
  • 119
  • $64$ is not close to the answer for the first one, indeed $x\approx 37.34$ That one can be "solved" using the Lambert W Function . The second clearly has $u=1$ as a solution (the only real solution, I believe). – lulu Jul 13 '22 at 17:58
  • Post edit: for the revised question, $x=64$ is now an exact solution, no need to apporximate. – lulu Jul 13 '22 at 18:05
  • Thank you, lulu. I made a mistake on (1). It is now written correctly. I actually did use Lambert on 1. Was looking to see if there was any other way. Thanks for your help. – Frederick Jul 13 '22 at 18:06

1 Answers1

2

  1. The functions $\ h_1(x)=x$ and $\ h_2(x)=$log$_8(x)-1$ are both strictly increasing and $h_1$ is positive on $\mathbb{R}_{>0}$, and thus so is the function given by their product: $$f(x)=x(\text{log}_8(x)-1)$$ Therefore there can be at most one solution to $f(x)=0$. Hence $x=64$ is the only solution to $f(x)=64$. (A strictly increasing function is injective!).

  2. If you study the sign of the derivative of $g(x)=x(x-1)-\text{log}(x)$, you can see that $g(x)$ has a global minimum at $x=1$, and moreover $g(1)=0$. Hence the only solution to $u(u-1)=\text{log}(u)$ is $u=1$.

Alessandro
  • 1,334