SOLUTION: log 6 +log (x-3)=2 log y 2y-x=3

SOLUTION: log 6 +log (x-3)=2 log y 2y-x=3

Algebra.Com

Question 422024: log 6 +log (x-3)=2 log y
2y-x=3
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
log 6 +log (x-3)=2 log y
10^(log 6)*10^(log (x-3))= 10^log y^2
6(x-3)=y^2
y^2=6x-18
2y-x=3
2y=x+3
y=(x+3)/2
(x^2+6x+9)/4=6x-18
x^2+6x+9=24x-72
x^2-18x+81=0
(x-9)^2=0
x=9
y=6
.
Ed

RELATED QUESTIONS
{{{log(2,x)+log(6,x)=3}}} (answered by lwsshak3)

log (x^2)(y) - .5 log x + 3 log... (answered by ccs2011)

log (x + 2) - 3 log... (answered by stanbon)

log(X+3)-log X=log... (answered by Fombitz)

Given Log(x)=3, Log(y)=4 and Log(z)=10, Find... (answered by stanbon)

solve, (1/2)Log(x^2y^4) + (2/3)Log(x^2y^6) to single a log with coefficient 1. (answered by lwsshak3)

How do I work log(x-1)-log(x+6)=log(x-2)-log(x+3) (answered by josgarithmetic)

2 log(x+6) + 1/3 log y - 5 log... (answered by Fombitz)

log (6x + 5) - log 3 = log 2 - log x (answered by rapaljer,arunpaul)