1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

We need

  • $x>0$

  • $\cos(\log x)>0$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.