SOLUTION: This equation is giving me issues. log(base2)(x-6)+log(base2)(x-4)-log(base2)x = 2 Solve for x. Reject any value of x that produces the logarithm of a negative number or 0. Wh

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: This equation is giving me issues. log(base2)(x-6)+log(base2)(x-4)-log(base2)x = 2 Solve for x. Reject any value of x that produces the logarithm of a negative number or 0. Wh      Log On


   

Question 5994: This equation is giving me issues.
log(base2)(x-6)+log(base2)(x-4)-log(base2)x = 2
Solve for x. Reject any value of x that produces the logarithm of a negative number or 0.
What I've done:
log(base2)(x-6)(x-4)-log(base2)x = 2
log(base2)(x-6)(x-4)/x = 2
(x-6)(x-4)/x = 2^2 = 4
4 = (x-6)(x-4)/x
There is where I am stuck.
Some help on this would be great. Thanks.
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
So I take it you're stuck at +4+=+%28%28x-6%29%28x-4%29%29%2Fx+.

All you'll do is cross-multiply. It might be easier if we put it this way:

+4%2F1+=+%28%28x-6%29%28x-4%29%29%2Fx+

Then we'll get +4x+=+%28x-6%29%28x-4%29+

+%28x-6%29%28x-4%29+-+4x+=+0+ <----- We switched the places of the expressions on both sides of the equals sign, and then moved the 4x to the other side so that everything equals zero.

+x%5E2+-+10x+%2B+24+-+4x+=+0+ <----- We expanded by using the product by using FOIL.

+x%5E2+-14x+%2B+24+=+0+ <--- Combined like terms

+%28x+-+12%29%28x+-+2%29+=+0+ <---- Refactored ("UnFoiled")

From looking at your above equation, x = 12 or x = 2.