2021/019) If $A = \log_{12} 18$ and $B = \log_{(24)}54$ then evaluate, AB + 5(A - B)?

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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