log2(x+1) = 5 - log2(x-3)
2006-12-03 17:06:58 · 2 answers · asked by thatonegirlamanda 2 in Science & Mathematics ➔ Mathematics
OMFG thanks i was putting 5^2 instead of 2^5
2006-12-03 17:22:00 · update #1
log2(x+1) = 5 - log2(x-3)
log2(x+1) + log2(x-3)= 5
log2((x + 1)(x - 3)) = 5
(x + 1)(x - 3) = 2^5
(x + 1)(x - 3) = 32
x^2 - 3x + 1x - 3 = 32
x^2 - 2x = 35
x^2 - 2x - 35 = 0
x = (2 +/- sqrt(4 - (4)(-35))) / 2
x = (2 +/- sqrt(144)) / 2
x = (2 +/- 12) / 2
x = 7 or -5
Check answers with original equation to make sure log of positive.
x = 7
2006-12-03 17:17:13 · answer #1 · answered by Michael M 6 · 1⤊ 0⤋
log base2 (x+1) = 5 - log base2 (x-3)
Move logs both to left.
log base2 (x+1) + log base2 (x-3) = 5
Added logs indicate multiplying expressions together.
log base2 (x+1)(x-3) = 5
Change into exponential notation
(x+1)(x-3) = 2^5
(x+1)(x-3) = 32
x^2 - 2x -3 = 32
x^2 - 3x - 35 = 0
(x - 7)( x + 5) = 0
These solve to x = 7 or x = -5, but since you can only take logs of positive numbers, you exclude -5.
The answer is 7.
2006-12-04 01:14:57 · answer #2 · answered by Pi R Squared 7 · 0⤊ 0⤋