# Precalc

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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 -

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 -

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 -

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 -

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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