Larger of $\cos(\ln x)$ or $\ln(\cos x)$

Question

Which is larger : $\cos(\ln x)$ or $\ln(\cos x)$ if $e^{-\pi/2} < x < \pi/2$

How do I solve such a problem? Any help would be welcome.

Answer 1

Remark that $\ln (\cos x)\leq 0$ and use the fact that $-\frac{\pi}{2}\leq \ln x\leq \ln\frac{\pi}{2}\leq \frac{\pi}{2}$ (in your domain), so $\cos(\ln x)\geq 0$ then $\ln(\cos x)\leq 0\leq \cos(\ln x)$.

Answer 2

$-1\leq \cos x \leq 1$, so $\ln(\cos x)\leq 0$.

$0 \le \cos x \le 1$ for $x \in [\frac{-\pi}{2},\frac{\pi}{2}]$, and since $(\ln e^{-\pi/2}, \ln \frac{\pi}{2})\subset [\frac{-\pi}{2},\frac{\pi}{2}]$, $0 < \cos x \le 1$ in our interval. Therefore

Hence, $\ln(\cos x) \le \cos(\ln x)$ for $x \in (e^{-\pi/2}, \frac{\pi}{2})$.