SOLUTION: log2(-x)+log2(x+12)=5 check for extraneous solutions. I got -8 and -4 I don't know if it's right and how many solutions.

Question 388779: log2(-x)+log2(x+12)=5 check for extraneous solutions. I got -8 and -4 I don't know if it's right and how many solutions.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
log2(-x)+log2(x+12)=5 check for extraneous solutions. I got -8 and -4 I don't know if it's right and how many solutions.
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log2(-x)+log2(x+12)=5
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log2[-x(x+12)] = 5
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-x(x+12) = 2^5
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-x^2-12x-32 = 0
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x^2+12x+32 = 0
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Factor:
(x+4)(x+8) = 0
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x = -4 or x = -8
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No extraneous zeros.
Cheers,
Stan H.
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