f(x) = log2x- -------> y=log2x-------- > 2^(y)=x (I)
therefore
in the equation 2^(2y) = 6---- >(2^(y))^2 = 6 (II)
(I) in (II)
((x))^2 = 6-----------------x=2.45
using a graphic tool
see graphic attached
for x=2.45 the value of y approximate is 1.2928
