1. Given $a, b>0, a \neq 1$, and $a^{b}=\log _{a} b$, then the value of $a^{a^{b}}-\log _{a} \log _{a} b^{a}$ is
1. -1 Explanation: From $a^{b}=\log _{a} b$ we know: $b=a^{a^{b}}, b=\log _{a} \log _{a} b$,
$$
a^{a^{b}}-\log _{a} \log _{a} b^{a}=b-\log _{a}\left(a \log _{a} b\right)=b-\log _{a} a-\log _{a} \log _{a} b=b-1-b=-1
$$