Question 389428: log2(2x+3)+log2x=1
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! log2(2x+3)+log2x=1
assuming base 2 on the logs
log2 (2x + 3) + log2 (x) = 1
logarithmic rule: logb (m) + logb (n) = logb (mn)
log2 x(2x + 3) = 1
logarithmic rule: if logb (y) = x then b^x = y
2^1 = 2 = x(2x + 3) = 2x^2 + 3x
0 = 2x^2 + 3x - 2
0 = (2x - 1)(x + 2) by FOIL --> 2x^2 - x + 4x - 2, same as above
2x - 1 = 0 --> 2x = 1 --> x = 1/2 or x + 2 = 0 --> x = -2
check x = 1/2:
log2 (2 * 1/2 + 3) + log2 (1/2) = 1
log2 (1 + 3) + log2 (1/2) = 1
log2 (4) + log2 (1/2) = log2 (2) = 1, yes
check x = -2:
log2 (2 * -2 + 3) + log2 (-2) = 1
log2 (-4 + 3) + log2 (-2) = 1
log2 (-1) + log2 (-2) = 1, no can not take logs of negative numbers
solution is only x = 1/2
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