Posted by Rebekah on .
1. Solve log (4x) = log (2) + log (x1)
the fact that this is x1 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 4x4) = 8
thanks so much!!! I have been studying and my final is tomorrow. these are just two questions that came up:)

Precalc 
Damon,
log [ 4x/(2(x1))] = 0
so
10^log [ 4x/(2(x1))] = 10^0 = 1
so
4x/[2(x1))] = 1
4x = 2 x  2
2 x =  2
x = 1 
Precalc 
Damon,
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
(x2)(x2) = 0
x = 2 
Precalc 
Reiny,
1. I disagree with Damon's solution.
log(4x) = log2 + log(x1) , so x>1
log(4x)  log(x1) = log2
log( 4x/(x1) ) = log2
4x/(x1) = 2
4x = 2x2
2x = 2
x = 1 , but x > 1 or else log (4x) and log(x1) are both undefined
no solution 
Precalc 
Rebekah,
thanks so much both of you!! Yah I wasn't sure about the first one but now I know how to do both!!
thanks!!