解方程log2(4^x+1)=x+log2(2^(x+3)-6)

log2(4^x+1)=x+log2[2^(x+3)-6]移项log2(4^x+1)-log2[2^(x+3)-6]=xlog2[(4^x+1)/(2^x×8-6)]=x即2^x=(4^x+1)/(2^x×8-6)去分母2^x(2^x×8-6）=4^x+18×4^x-6×2^x=4^x+17×4^x-6×2^x-1=0令2^x=t,则t＞07t²-6t-...