SOLUTION: Ok, I'd like to say thank you in advance. The problem is: log(base 2)(6-x)+ log(base 2)(2-x)=5. I'm not sure if I went about it right but I tried multiplying and got: log(base 2)(x

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Ok, I'd like to say thank you in advance. The problem is: log(base 2)(6-x)+ log(base 2)(2-x)=5. I'm not sure if I went about it right but I tried multiplying and got: log(base 2)(x      Log On


   

Question 337908: Ok, I'd like to say thank you in advance. The problem is: log(base 2)(6-x)+ log(base 2)(2-x)=5. I'm not sure if I went about it right but I tried multiplying and got: log(base 2)(x^2 -8x+12)=5. Please help.
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
good so far

x^2 - 8x + 12 = 2^5 ___ x^2 - 8x + 12 = 32

x^2 - 8x - 20 = 0

(x + 2)(x - 10) = 0


REMEMBER : logarithms are NOT defined for negative quantities
___ check your answers by plugging back in to the original equation