Find domain of function $f(x)=\log\left(\cos\left(\log x\right)\right)$

At the beginning $x>0$
but I have no idea how to handle logarithm
$\cos\left(\log x\right)>0\\\log x>\arccos 0=\frac{\pi}{2}\\\log x>\log 10^{\frac{\pi}{2}}\Rightarrow x>10^{\frac{\pi}{2}}$
I think I'm doing something wrong


We need

  • $x>0$

  • $\cos(\log x)>0$

$$\implies 0\le \log x < \frac \pi 2 \cup -\frac \pi 2 +2n\pi < \log x <\frac \pi 2 +2n\pi$$

  • $\begingroup$ so the solution will be $-\frac{\pi}{2}+2\pi n<\log x<\frac{\pi}{2}+2\pi n,n\in\mathbb{Z}\\\log10^{-\frac{\pi}{2}+2\pi n}<\log x<\log 10^{\frac{\pi}{2}+2\pi n}\Rightarrow\\x\in\left(10^{-\frac{\pi}{2}+2\pi n};10^{\frac{\pi}{2}+2\pi n}\right)$ $\endgroup$ – vmahth1 Nov 27 '19 at 7:11
  • 1
    $\begingroup$ @user On a lighter note, you were gimusi, weren't you? Why did you change your name? A change in philosophy? "gimusi" is born in Lithuanian, by the way. $\endgroup$ – астон вілла олоф мэллбэрг Nov 27 '19 at 7:21
  • $\begingroup$ yes i'm sure in my book $\log x=\log_{10}x\\\ln x=\log_ex$ $\endgroup$ – vmahth1 Nov 27 '19 at 7:26
  • $\begingroup$ @астонвіллаолофмэллбэрг Yes I was! Just a change, nothing else. Bye $\endgroup$ – user Nov 27 '19 at 15:50
  • 2
    $\begingroup$ @vmahth1 Yes that's fine for the solutions $-\frac \pi 2 +2n\pi < \log x <\frac \pi 2 +2n\pi$ with $n\ge 1$. Don't forget also $0\le \log x < \frac \pi 2$. $\endgroup$ – user Nov 27 '19 at 15:52

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