Algebraic expression for the period of $\cos (\log (x))$? This question relates only to $x \in \mathbb{R}^+$. The function $f(x) = \cos (\log (x))$ is clearly defined on the positive reals, with a monotonic decreasing period $p(x)$ which is defined at the limits of this range by
$$\underset{x\to 0^+}{\text{lim}}p(f(x)) = \infty$$
$$\underset{x\to \infty}{\text{lim}}p(f(x)) = 0$$
$p$ still needs clear definition for any given $0 < x_0 < \infty$. So, let us define $p(x_ 0)$ as the smallest positive real $a$ for which this pair of equations hold:
$$\cos (\log (x_0)) = \cos (a \log (x_0))$$
$$\text{sgn}\left(\frac{\partial \cos \left(\log \left(x_0\right)\right)}{\partial x_0}\right) = \text{sgn}\left(\frac{\partial \cos \left(a \log \left(x_0\right)\right)}{\partial  x_0}\right)$$
Is the right approach? And how do I proceed?
 A: I would approach this as follows.
The period of $\cos(x)$ is $2\pi$, everywhere. At a particular $x_0$, what $a$ must you add to $x_0$ so that $\ln(x_0+a)$ is $2\pi$ greater than $\ln(x_0)$? In other words, solve for $a$ in
$$\ln(x_0)+2\pi=\ln(x_0+a)$$
The solution is
$$a=\left(e^{2\pi}-1\right)x_0$$

It would also make sense to me to go backwards and solve
$$\ln(x_0)-2\pi=\ln(x_0-a)$$
where the solution is
$$a=\left(1-e^{-2\pi}\right)x_0$$
You could take the two solutions and average them somehow. It might make sense to take their geometric mean:
$$a=\left(e^{\pi}-e^{-\pi}\right)x_0$$
A: $\cos x = \cos y\iff y = \pm x + 2k \pi$ for some integer $k$
$\log w = \log u \iff w = u$ so
$\cos (\log x) = \cos (\log y) \iff \log x = \pm \log y + 2k \pi $
$\iff \log x \mp \log y = 2k\pi\iff \log (\begin{cases}xy\\\frac xy\end{cases}) = 2k\pi$
And if there were a period, $p > 0$, so that $\cos (\log (x + p))=\cos (\log x)$ for all $x$ we would have that for every $x$ that either $\log (x(x+p))$ or $\log \frac x{x+p} = 2k\pi$.
But $\log(x(x+p))$ and $\log\frac x{x+p}$ are both continuous and not constant they can't be restricted to only multiples of $2\pi$, $\cos (\log x)$ is not periodic.
