Question
Solve this simultaneously log(x-2) + log 2 = 2log y and log( x -3y+3) = 0
Answers
oobleck
log(x-2) + log 2 = 2log y
log( x -3y+3) = 0
removing the logs, this becomes
2(x-2) = y^2
x-3y+3 = 1
since x = 3y-2, we now have
2(3y-2)-4 = y^2
y^2-6y+8 = 0
(y-2)(y-4) = 0
y = 2 or 4
so x = 4 or 10
log( x -3y+3) = 0
removing the logs, this becomes
2(x-2) = y^2
x-3y+3 = 1
since x = 3y-2, we now have
2(3y-2)-4 = y^2
y^2-6y+8 = 0
(y-2)(y-4) = 0
y = 2 or 4
so x = 4 or 10