Given that log_{x} (7y+1) - log_{x} (2y) =1 x>4, 0<y<1 , express y in terms of x.

log_{x} (7y+1) - log{x} (2y) =1 --> log_{x} [(7y+1)/2y]=1 (y =/= 0, Rules of logarithms i.e. difference of logarithms) --> x = [(7y+1)/2y] (x>0, Rules of logarithms i.e. log_{x} x = 1) --> 2yx = 7y+1 (Multiply by 2y) --> 2yx-7y= 1 (Moving y's to one side) --> y(2x-7) = 1 (Factorising out the y) --> y = 1/(2x-7) 

Answered by Christopher L. Maths tutor

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