The value of 6 + log3/2(1/(3√2 - √4)) is

Let y = √(4/(3√2 - √4)).
Hence y² = 4/(3√2 - √4)
y³√2 = y² + y√2 = 0
y = (-√2 ± √(2 + 4(72)))/6√2
y = (-√2 ± √176)/6√2
Since input of logarithmic functions cannot be negative. Therefore y = (176√2)/6√2 = 4√2/3
Hence 1/(3√2 - √4) = 1/(3√2 - 2) = 4√2/3
Therefore 6 + log3/2(4√2/3) = 6 + log3/2((3/2)²) = 6 + 2 = 4