You can put this solution on YOUR website! Apply log rules. Review them at:
http://www.purplemath.com/modules/logrules.htm
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Log(2) (x-8) + Log(2) (x+4) = 5
Log(2) [(x-8)(x+4)] = 5
(x-8)(x+4) = 2^5
(x-8)(x+4) = 32
x^2-4x-32 = 32
x^2-4x-64 = 0
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Since we can't factor, we must use the quadratic formula. Doing so yields:
x = {10.246, -6.246}
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Details of quadratic below:
You can put this solution on YOUR website! Log(2) (x-8) + Log(2) (x+4) = 5
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Assuming you mean log base 2: Adding logs implies multiplying
(x-8)*(x+4)) = 2^5 = 32