1. Let set $A=\left\{a-1,2 \log _{2} b\right\}$ and $B=\left\{a+1, \log _{2}(16 b-64)\right\}$ have exactly one common element which is $a$, then the real number $a=$
Ronggou 6.
Analysis: Since $a-1 \neq a, a+1 \neq a$, the common element is $2 \log _{2} b=\log _{2}(16 b-64)$, solving this gives $b=8, a=2 \log _{2} b=6$.