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

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Answer 1 · 2016-04-07T13:28:29

$y = {- \frac{1}{5}, 7}$ $y={-1/5,7}$

${log (2 y + 4)}^{2} = log 9 + {log (y - 1)}^{2}$ $log(2y+4)^2=log9+log(y-1)^2$

${log (2 y + 4)}^{2} = log [9 {(y - 1)}^{2}]$ $log(2y+4)^2=log[9(y-1)^2]$

${(2 y + 4)}^{2} = 9 {(y - 1)}^{2}$ $(2y+4)^2=9(y-1)^2$

$4 y^{2} + 16 y + 16 = 9 (y^{2} - 2 y + 1)$ $4y^2+16y+16=9(y^2-2y+1)$

$4 y^{2} + 16 y + 16 = 9 y^{2} - 18 y + 9$ $4y^2+16y+16=9y^2-18y+9$

$9 y^{2} - 18 y + 9 - 4 y^{2} - 16 y - 16 = 0$ $9y^2-18y+9-4y^2-16y-16=0$

$5 y^{2} - 34 y - 7 = 0$ $5y^2-34y-7=0$

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

$Delta=36$

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

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

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

How do you solve 2log (2y + 4) = log 9 + 2log (y -1)2log(2y+4)=log9+2log(y−1)2log (2y + 4) = log 9 + 2log (y -1)?