Asked by Sara
log(subscript)2 (x+5) + log(subscript)2 (x-1) =4 find the extraneous root
Answers
Answered by
Reiny
log<sub>2</sub> (x+5) + log<sub>2</sub> (x-1) =4 , x > 1 or else the log .... would be undefined
log<sub>2</sub> ( (x+5)(x-1) ) = 4
(x+5)(x-1) = 2^4
x^2 + 4x - 5 - 16 = 0
x^2 + 4x - 21 = 0
(x + 7)(x - 3) = 0
x = -7 or x = 3
So x = 3, the other root is extraneous
log<sub>2</sub> ( (x+5)(x-1) ) = 4
(x+5)(x-1) = 2^4
x^2 + 4x - 5 - 16 = 0
x^2 + 4x - 21 = 0
(x + 7)(x - 3) = 0
x = -7 or x = 3
So x = 3, the other root is extraneous
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