Asked by alejandro
Solve log2(x-1) = 5-log2(x+3) for x.
base 2.
base 2.
Answers
Answered by
Henry
Log2(x-1) = 5 - Log2(x+3).
Log2(x-1) + Log2(x+3) = 5
Log2((x-1)(x+3)) = 5
Log2(x^2+3x-x-3) = 5
Log2(x^2+2x-3) = 5
x^2 + 2x -3 = 2^5 = 32
x^2 + 2x -3 -32 = 0
x^2 + 2x -35 = 0
(x-5)(x+7) = 0
x-5 = 0
X = 5.
x+7 = 0
X = -7.
Solution set: X = -7, X = 5.
Log2(x-1) + Log2(x+3) = 5
Log2((x-1)(x+3)) = 5
Log2(x^2+3x-x-3) = 5
Log2(x^2+2x-3) = 5
x^2 + 2x -3 = 2^5 = 32
x^2 + 2x -3 -32 = 0
x^2 + 2x -35 = 0
(x-5)(x+7) = 0
x-5 = 0
X = 5.
x+7 = 0
X = -7.
Solution set: X = -7, X = 5.
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