Question 410462: please help me solve these equations
3log(3y+2) = log(3y+2) - log7
and
( log(5x-6) / logx ) = 2
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 3log(3y+2) = log(3y+2) - log7
and( log(5x-6) / logx ) = 2
..
solving second equation,
log(5x-6)/logx=2
=log(5x-6)=2logx
log(5x-6)-2logx=0
log(5x-6)/x^2=0
convert to exponential form,
(5x-6)/x^2=10^0=1
5x-6=x^2
x^2-5x+6=0
(x-3)(x-2)=0
x-3=0
x=3
x-2=0
x=2
ans:
x=2
x=3
..
solving first equation
3log(3y+2)=log(3y+2)-log7
3log(3y+2)-log(3y+2)+log7=0
2log(3y+2)+log7=0
log(3y+2)^2*7)=0
convert to exponential form,
7(3y+2)^2=10^0=1
7(9y^2+12y+4)=1
63y^2+84y+28=1
63y^2+84y+27=0
divide by 3
21y^2+28y+9=0
solve by quadratic formula,
a=21,b=28,c=9
y=[-28+-sqrt(28^2-4*21*9)]/2*21
=[-28+-sqrt(28)]/42
=(-28+-5.29)/42
y=-.792 (reject because this makes (3y+2)<0) In logx,x>0
or
y=-.541
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