SOLUTION: I am having trouble solving equations with logs this is the problem i am stuck on... log6(x+3)+log6(x+4)=1

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Question 111220: I am having trouble solving equations with logs this is the problem i am stuck on...
log6(x+3)+log6(x+4)=1
Answer by josmiceli(6784) About Me  (Show Source):
You can put this solution on YOUR website!
log%286%2C%28x%2B3%29%29%2Blog%286%2C%28x%2B4%29%29+=+1
log%286%2C%28x%2B3%29%28x%2B4%29%29+=+1
Written the other way,
%28x%2B3%29%28x%2B4%29+=+6%5E1
x%5E2+%2B+7x+%2B+12+=+6
subtract 12 from both sides and complete the square
x%5E2+%2B+7x+%2B+%2849%2F4%29+=+-6+%2B+%2849%2F4%29
%28x+%2B+%287%2F2%29%29%5E2+=+%2849%2F4%29+-+%2824%2F4%29
%28x+%2B+%287%2F2%29%29%5E2+=+25%2F4
take square root of both sides
x+%2B+7%2F2+=+5%2F2
x+=+-1 answer
Now plug into original equation to check
log%286%2C%28x%2B3%29%29%2Blog%286%2C%28x%2B4%29%29+=+1
log%286%2C+-1+%2B+3%29+%2B+log%286%2C+-1+%2B+4%29+=+1
log%286%2C+2%29+%2B+log%286%2C+3%29+=+1
log%286%2C+2%2A3%29+=+1
log%286%2C+6%29+=+1 checks OK
rewite it
6%5E1+=+6 true