If #a=log_12 18# & #b = log_24 54#, wha tis the value of #ab+5(a-b)#?

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Answer 1 · 2017-05-25T23:14:24

One.

Work on basis 10.

#a = (log 18)/(log 12) = (log 2 + 2log3)/(log 3 + 2 log 2) and b = (log 2 + 3 log 3)/(log 3 + 3 log 2)#

#log 2 = x, log 3 = y#

#ab = (x + 2y)/(y + 2 x) * (x + 3 y)/(y + 3 x) = frac{x^2 + 5xy + 6y^2}{(y + 2x)(y + 3x)}#

#a - b = (x + 2y)/(y + 2 x) - (x + 3 y)/(y + 3 x) = frac{(x + 2y)(3x + y) - (x + 3y)(2x + y)}{(y + 2x)(y + 3x)}#

#= frac{3x^2 + xy + 6xy + 2y^2 - 2x^2 -xy -6xy - 3y^2}{(y + 2x)(y + 3x)}#

#ab + 5(a - b) = frac{x^2 + 5xy + 6y^2 + 5(x^2 - y^2)}{(y + 2x)(y + 3x)}#

#= frac{6x^2 + 5xy + y^2}{(y + 2x)(y + 3x)} = 1#