关于x的方程log2(x+√(x^2-1))-a=0有实数解,求实数a的取值范围,并求方程的解

log2(x+√(x²-1))-a=0log2(x+√(x²-1))=aa≥0,方程有解,所以a取值范围≥02^a=(x+√(x²-1))2^a-x=√(x-1)2^2a-2x2^a+x²=x²-12^2a-2x2^a=-1x=(2^2a+1)/2^(a+1)