SOLUTION: Please tell me if I am doing this right, the equation is {log6 (x+3) + log6 (x+4)= 1} I end up with x=-6 and x=-1 but they cant have a negative answer??

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please tell me if I am doing this right, the equation is {log6 (x+3) + log6 (x+4)= 1} I end up with x=-6 and x=-1 but they cant have a negative answer??      Log On


   

Question 296846: Please tell me if I am doing this right, the equation is {log6 (x+3) + log6 (x+4)= 1} I end up with x=-6 and x=-1 but they cant have a negative answer??
Found 2 solutions by richwmiller, Alan3354:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Toss the -6 but keep the -1. Why?

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Please tell me if I am doing this right, the equation is {log6 (x+3) + log6 (x+4)= 1} I end up with x=-6 and x=-1 but they cant have a negative answer??
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log%286%2C%28x%2B3%29%29+%2B+log%286%2C%28x%2B4%29%29+=+1
log%286%2C%28x%2B3%29%2A%28x%2B4%29%29+=+1
(x+3)*(x+4) = 6
x%5E2+%2B+7x+%2B+12+=+6
x%5E2+%2B+7x+%2B+6+=+0
(x+1)*(x+6) = 0
x = -6 would make it the log of a negative number, so ignore that one.
x = -1 works, because the arguments of the logs are both positive.
--> log%286%2C2%29+%2B+log%286%2C3%29+=+1