SOLUTION: log(x+1)+log(x-2)=1

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Question 193653: log(x+1)+log(x-2)=1
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
log(x+1)+log(x-2)=1
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log((x+1)*(x-2)) = 1
(x+1)*(x-2) = 10
x%5E2+-+x+-+2+=+10
x%5E2+-+x+-+12+=+0
(x-4)*(x+3) = 0
x = 4
x = -3

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In general:
log%28a%29+%2B+log%28b%29+=+log%28ab%29 so,
log%28%28x%2B1%29%29%2Blog%28%28x-2%29%29=+log%28%28x%2B1%29%2A%28x-2%29%29
also, 1+=+log%2810%29
log%28%28x%2B1%29%2A%28x-2%29%29+=+log%2810%29
If the logs are equal, the antilogs are equal
%28x%2B1%29%2A%28x-2%29+=+10
x%5E2+%2B+x+-+2x+-+2+-+10+=+0
x%5E2+-+x+-+12+=+0
%28x+-+4%29%2A%28x+%2B+3%29+=+0
x+=+4
x+=+-3 this result doesn't work, becase it
gives me a log%28-5%29 and no log can give a
negative answer
check:
x+=+4
log%28x%2B1%29%2Blog%28x-2%29=+1
log%285%29+%2B+log%282%29+=+1
.6989700043+%2B+.3010299957+=+1
.9999999997+=+1 (inaccuracy is due to rounding off)
OK