19.已知函数f(x)={2x−1,x⩾alog12(x+1),−1<x<af(x)=\left\{{\left.\begin{array}{l}{{2^x}-1,x\geqslant a}\\ {{{\log }_{\frac{1}{2}}}({x+1}),-1<x<a}\end{array}\right.}\right.f(x)={2x−1,x⩾alog21(x+1),−1<x<a,若函数g(x)=f(x)−2g(x)=f(x)-2g(x)=f(x)−2有两个零点,则实数aaa的取值范围是((( )))