Question 5994
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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 (2, (x - 6)) + log (2, (x - 4)) - log (2, (x)) = 2}}}

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

So, we get: {{{log (2, (x - 6)) + log (2, (x - 4)) - log (2, (x)) = 2}}}, <font color = red><font size = 4><b>with x > 6</font></font></b>

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: <font color = red><font size = 4><b>Reject any value of x that produces the logarithm of a negative number or 0.</font></font></b></pre>