The correct options are
A 14
C 13
Taking log on both sides of given equations,
log3log(3x)=log4log(4y)
log3(log3+logx)=log4(log4+logy)[∵log(ab)=loga+logb] ⇒ (i)
And, logxlog4=logylog3
logx(log4)=(logy)log3 ⇒ (ii)
From (i)
(log4)2−(log3)2=log3logx−log4logy ⇒ (iii)
From (ii)
logxlog3=logylog4=λ(say) ⇒ (iv)
From (iv) and (iii)
(log4)2−(log3)2=[(log3)2−(log4)2]λ
∴λ=−1
∴logx=−log3=log13
⇒logy=−log4=log14
∴x=13,y=14