Solve simultaneously log (x-1)+2logy=2log3 and logx+logy=log6

Answer 1

From the first, using log rules:
log (x-1)+2logy=2log3
log (x-1)+log(y^2)=log(3^2)
log( (x-1)(y^2) ) = log 9
(x-1)y^2 = 9

from the 2nd:
logx+logy=log6
log(xy) = log6
xy = 6 or x = 6/y

now use substitution,
(6/y - 1)(y^2) = 9
6y - y^2 - 9 = 0
y^2 - 6y + 9 = 0
(y - 3)^2 = 0
y - 3 = 0
y = 3
then x = 6/3 = 2

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