How do you solve #2log (2y + 4) = log 9 + 2log (y -1)#?

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Answer 1 · 2016-04-07T13:28:29

#y={-1/5,7}#

#log(2y+4)^2=log9+log(y-1)^2#

#log(2y+4)^2=log[9(y-1)^2]#

#(2y+4)^2=9(y-1)^2#

#4y^2+16y+16=9(y^2-2y+1)#

#4y^2+16y+16=9y^2-18y+9#

#9y^2-18y+9-4y^2-16y-16=0#

#5y^2-34y-7=0#

#Delta=sqrt((-34)^2+4*5*7)=sqrt(1156+140)=sqrt1296#

#Delta=36#

#y_1=(34-36)/(2*5)=-(2)/10=-1/5#

#y_2=(34+36)/(2*5)=70/10=7#

#y={-1/5,7}#