81 Given $0<x<\frac{\pi}{2}, \log _{2} \cos x$ and $\log \sqrt{\tan x}$ both have a characteristic of zero, and the sum of their mantissas is 1, then $x=$
$81 \operatorname{arc} \sin \frac{\sqrt{5}-1}{2}$. According to the problem,
$$
\log _{\sin x} \cos x+\log _{\cos x} \sqrt{\tan x}=1 .
$$

Let $\log _{\sin x} \cos x=a, 0<a<1$, then
$$
\begin{aligned}
\log _{\cos x} \sqrt{\tan x} & =\frac{\log _{\sin x} \sqrt{\tan x}}{\log _{\sin x} \cos x}=\frac{\frac{1}{2}\left(\log _{\sin x} \sin x-\log _{\sin x} \cos x\right)}{a} \\
& =\frac{1-a}{2 a},
\end{aligned}
$$

Substituting into (1) we have
$$
a+\frac{1-a}{2 a}=1,
$$

Solving for $a$ gives $a=\frac{1}{2}$. Therefore,
$$
\log _{\sin x} \cos x=\frac{1}{2},
$$

Thus we get
$$
\sin ^{2} x+\sin x-1=0,
$$

Hence
$$
\sin x=\frac{\sqrt{5}-1}{2},\left(0<x<\frac{\pi}{2}\right),
$$

Therefore
$$
x=\arcsin \frac{\sqrt{5}-1}{2} \text {. }
$$