SOLUTION: b. log_6(x+4)+log_6(x-1)=1

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Question 1076619: b. log_6(x+4)+log_6(x-1)=1
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log_6(x+4)+log_6(x-1)=1
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log6[(x+4)(x-1)] = 1
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(x+4)(x-1) = 6
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x^2 + 3x -10 = 0
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(x+5)(x-2) = 0
Positive solution::
x = 2
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

log%286%2C%28x%2B4%29%29%2Blog%286%2C%28x-1%29%29=1

log%286%2C%28x%2B4%29%28x-1%29%29=1 .....since log%286%2C6%29=1 we have

log%286%2Cx%5E2+%2B+3+x+-+4%29=log%286%2C6%29.....if log same, than

x%5E2+%2B+3x+-4=6

x%5E2+%2B+3x+-4-6=0

x%5E2+%2B+3x+-10=0....factor completely

x%5E2+%2B+5x-2x+-10=0

%28x%5E2+%2B+5x%29-%282x+%2B10%29=0

x%28x+%2B+5%29-2%28x+%2B5%29=0

%28x+-+2%29%28x+%2B+5%29+=+0

solutions:
if %28x+-+2%29+=+0=>x=2
if %28x+%2B+5%29+=+0=>x=-5......disregard negative solution
so, your solution is highlight%28x=2%29