How do you solve $2log_3y-log_3(y+4)=2$ ?

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Answer 1 · 2015-11-24T23:54:43

$y=12$

Recall that:

${(bloga=loga^b),(loga-logb=log(a/b)):}$

$2log_3y-log_3(y+4)=2$

$log_3(y^2)-log_3(y+4)=2$

$log_3(y^2/(y+4))=2$

$y^2/(y+4)=3^2=9$

$y^2=9(y+4)=9y+36$

$y^2-9y-36=0$

$(y-12)(y+3)=0$

$y=12$ or $color(red)cancel(y=-3$

(In $loga$ , $a>0$ .)

How do you solve 2log_3y-log_3(y+4)=22log_3y-log_3(y+4)=2?