If 2x+y=6y and 3x−1=2y+1 then the value of (log⁡3−log⁡2)/(x−y) is

Taking log, we have     (x+y)log⁡2=y(log⁡2+log⁡3)∴     xlog⁡2=ylog⁡3 or     xlog⁡3=ylog⁡2=x−ylog⁡3−log⁡2=λ (say) --ialso (x−1)log⁡3=(y+1)log⁡2now xlog⁡3−ylog⁡2=log⁡3+log⁡2Using Eq. (i), we getλ(log⁡3)2−(log⁡2)2=log⁡3+log⁡2λ=1log⁡3−log⁡2∴1λ=1log⁡3−log⁡2⇒λ=log⁡32