3. Let $a>1$. Then when the graphs of $y=a^{x}$ and $y=\log _{a} x$ are tangent, $\ln \ln a=$ $\qquad$

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3. -1.

Since the two functions are inverse functions of each other and are symmetric about the line $y=x$, the point of tangency lies on $y=x$.
Let the point of tangency be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
x_{0}=a^{x_{0}}, \\
a^{x_{0}} \ln a=1 .
\end{array}
$$

Substituting equation (1) into equation (2) gives $x_{0} \ln a=1$, i.e.,
$$
\ln a^{x_{0}}=1 \text{. }
$$

Substituting equation (1) into equation (3) gives
$$
\ln x_{0}=1 \Rightarrow x_{0}=e \text{. }
$$

Thus, $\ln a=\frac{1}{\mathrm{e}} \Rightarrow \ln \ln a=-1$.