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

so

10^log [ 4x/(2(x-1))] = 10^0 = 1

so

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

solve quadratic

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

no solution

- 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!!

thanks!!

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