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.

Found 2 solutions by prince_abubu, MathTherapy:
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.

Answer by MathTherapy(10699) About Me  (Show Source):
You can put this solution on YOUR website!
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.

The person who responded is PARTIALLY WRONG!!
log+%282%2C+%28x+-+6%29%29+%2B+log+%282%2C+%28x+-+4%29%29+-+log+%282%2C+%28x%29%29+=+2

The smallest of the 3 logs is x - 6, so x - 6 MUST be > 0, and so, x > 6.

So, we get: log+%282%2C+%28x+-+6%29%29+%2B+log+%282%2C+%28x+-+4%29%29+-+log+%282%2C+%28x%29%29+=+2, with x > 6

The person who responded has solutions, x = 12 and x = 2, but as you can see, the x value, 12 is > 6, but the
other value 2, is NOT, thereby making the solution, x = 12, ACCEPTABLE, and x = 2, EXTRANEOUS, and UNACCEPTABLE!

The person must've missed this: Reject any value of x that produces the logarithm of a negative number or 0.