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) (Show Source): Answer by MathTherapy(10699) (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!!
The smallest of the 3 logs is x - 6, so x - 6 MUST be > 0, and so, x > 6.
So, we get: , 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.
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