Posted by Rebekah on Thursday, December 20, 2012 at 8:18pm.
1. Solve log (4x) = log (2) + log (x-1)
the fact that this is x-1 and not + is really messing me up because i keep getting a negative number and that isnt possible
2. Solve: absolut value of(x^2 -4x-4) = 8
thanks so much!!! I have been studying and my final is tomorrow. these are just two questions that came up:)
Precalc - Damon, Thursday, December 20, 2012 at 10:02pm
log [ 4x/(2(x-1))] = 0
10^log [ 4x/(2(x-1))] = 10^0 = 1
4x/[2(x-1))] = 1
4x = 2 x - 2
2 x = - 2
x = -1
Precalc - Damon, Thursday, December 20, 2012 at 10:06pm
x^2 - 4 x - 4 = 8
x^2 - 4 x - 12 = 0
then other possible solution
x^2 - 4 x -4 = -8
x^2 - 4 x + 4 = 0
(x-2)(x-2) = 0
x = 2
Precalc - Reiny, Thursday, December 20, 2012 at 11:54pm
1. I disagree with Damon's solution.
log(4x) = log2 + log(x-1) , so x>1
log(4x) - log(x-1) = log2
log( 4x/(x-1) ) = log2
4x/(x-1) = 2
4x = 2x-2
2x = -2
x = -1 , but x > 1 or else log (4x) and log(x-1) are both undefined
Precalc - Rebekah, Friday, December 21, 2012 at 7:04am
thanks so much both of you!! Yah I wasn't sure about the first one but now I know how to do both!!
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