已知关于x的方程2x^2+(log2^m)x+log2^根号m=0有两个相等的实数根,求m的值

有两个相等的实数根
判别式等于0
所以(log2^m)^2+8log2^√m=0
(log2^m)^2+4*2log2^√m=0
(log2^m)^2+4log2^(√m)^2=0
(log2^m)^2+4log2^m=0
log2^m(log2^m+4)=0
log2^m=0,log2^m=-4
m=1,m=2^(-4)=1/16