Question 401278: log8^(n-3)+log8(n+4)=1
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! log8(n-3)+log8(n+4)=1
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log8[n^2+n-12] = 1
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n^2+n-12 = 8
n^2+n-20 = 0
(n+5)(n-4) = 0
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n cannot be -5 because log8(-8) is meaningless.
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n = 4 gives log8(4-3)+log8(4+4) = 1
gives log8(1) + log8(8) = 1
so 0+1 = 1
1 = 1
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Solution: n=4
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Cheers,
Stan H.
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