We have
A = \frac{\log18}{\log12} = \frac{log\,2 + 2\log\,3}{2\log\, 2 + \log\, 3}
A = \frac{\log 54}{\log 24} = \frac{log\,2 + 3\log\,3}{3\log\, 2 + \log\, 3}
putting x=\log\, 2 and y=\log\,3 we get
A= \frac{x+2y}{2x+y}
and B= \frac{x+3y}{3x+y}
So
AB + 5(A - B)
= \frac{(x+2y)(x+3y)}{(2x+y)(3x+y)} + 5(\frac{(x+2y)(3x+y) - (2x+y)(x+3y)}{ (2x+y)(3x+y)})
= \frac{[(x^2+5xy+6y^2) + 5(x^2 - y^2)]}{ (2x+y)(3x+y)}
= \frac{6x^2 + 5xy + y^2}{x^2 + 5xy + y^2} =1
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