13. Given constants $a, b$ satisfy $a, b>0, a \neq 1$, and points $P(a, b), Q(b, a)$ are both on the curve $y=\cos (x+c)$, where $c$ is a constant. Then $\log _{a} b=$ $\qquad$
13. 1 .

Given points $P(a, b), Q(b, a)$ are both on the curve
$$
y=\cos (x+c)
$$

we know
$$
\begin{array}{l}
a-b=\cos (b+c)-\cos (a+c) \\
=2 \sin \left(\frac{a+b+2 c}{2}\right) \cdot \sin \frac{a-b}{2} .
\end{array}
$$

Without loss of generality, assume $a \geqslant b$.
If $a>b$, then
$$
\left|\sin \frac{a+b+2 c}{2}\right| \leqslant 1,\left|\sin \frac{a-b}{2}\right|<\frac{a-b}{2} \text {. }
$$

Thus $|a-b|<2\left|\frac{a-b}{2}\right|=|a-b|$, which is a contradiction.
Since the graphs of $y=x$ and $y=\cos (x+c)$ must intersect, we know that $a=b$ holds.
Therefore, $\log _{a} b=1$.