SOLUTION: Solve log5(x-2)+log5(x+2)<1

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Question 1200726: Solve log5(x-2)+log5(x+2)<1
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39847) About Me  (Show Source):
You can put this solution on YOUR website!
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Solve log5(x-2)+log5(x+2)<1
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??

log%285%2C%28x-2%29%29%2Blog%285%2C%28x%2B2%29%29%3C1
?

log%285%2C%28x-2%29%28x%2B2%29%29%3C1
log%285%2C%28x-2%29%28x%2B2%29%29%3Clog%285%2C5%29
.
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Answer by MathTherapy(10895) About Me  (Show Source):
You can put this solution on YOUR website!
Solve log5(x-2)+log5(x+2)<1
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The smaller of the 2 log-arguments, x - 2, MUST be > 0. So, x - 2 > 0___x > 2. We now have:
log+%285%2C+%28x+-+2%29%29+%2B+log+%285%2C+%28x+%2B+2%29%29+%3C+1, with x > 2.
       log+%285%2C+%28%28x+-+2%29%28x+%2B+2%29%29+%3C+1%29 --- Applying log+%28b%2C+%28a%29%29+%2B+log+%28b%2C+%28c%29%29+=+log+%28b%2C+%28ac%29%29
                 %28x+-+2%29%28x+%2B+2%29+%3C+5%5E1 -- Converting to EXPONENTIAL form
                             x%5E2+-+4+%3C+5
                             x%5E2+-+9+%3C+0
                (x + 3)(x - 3) < 0
The above factors DENOTE 2 CRITICAL points for x: - 3, and 3. 

With CRITICAL x-points being - 3 and 3, the solution-set would otherwise be in the interval between
these 2 x-points, non-inclusive, but for the fact that x > 2, as stated above. So, critical x-points now
become 2 and 3, with the solution set between these 2 points, non-inclusive.
This gives us: 2 < x < 3, or in interval notation, ()