college algebra

Question

find the domain of the logarithmic function

 

f(x)=log(x+6/x-5)

 

 

Answer 1

OR

We can be BRAVE, give the customers what they want,

and leave the argument of the log as is and STILL find the domain

rather than ASSUME there are parenthesis. Albeit, the logic gets rather interesting.

 

 

X + 6/x - 5 =

x^2 + 6 - 5x
------------ =
        x

x^2 - 5x + 6
------------ =
       x


(x - 2)(x - 3)
--------------
         x

For this algebraic fraction to be positive,
both numerator and denominator MUST have the
same signs.

x>0 AND (x-2)(x-3)>0

 

                (x-2)>0 and (x-3)>0       or          x-2<0 and x-3<0
                 x>2 and x>3 --> x>3    or          x<2 and x<3--->x<2

 

x>0 and x>3   or   x>0 and x<2

     x>3   or     0 < x<=2

0< x<=2 or x>=3

------------------------------------------------

 

Now it will be proven by contradiction that

x must be positive in the context of this algebraic fraction.

 

Suppose x<0
In order for the algebraic fraction to be positive, (x-2)(x-3)<0

                x-2>0 and x-3<0   or     x-2<0 and x-3>0

                 x>2 and x<3         or         x<2 and x>3

As the latter case is a contradiction, 2 < x < 3

which contradicts the assumption that x<0.          

 

------------------------------------------------------------------

 

The argument in and of itself is a function, whose graph

shows the domain highlighted in BOLD above

                    0< x<=2 or x>=3

 

But there's the method and logic behind the madness.

 

 
     

Answer 2

I am assuming that you left out parentheses and that the function is f(x) = log[(x+6)/(x-5)].

 

Since we can't find the logarithm of 0 or the logarithm of a negative number, we must have (x+6)/(x-5) > 0.

 

There are 2 cases to consider:

 

  Case 1:  x+6>0 and x-5>0

 

                Then x>-6 and x>5.   So, x>5.

 

  Case 2:  x+6<0 and x-5<0

 

                Then x<-6 and x<5.  So, x<-6.

 

Domain:  (-∞, -6) ∪ (5,∞)

 

Answer 3

You mean that the argument of the log function is (x+6)/(x-5). These extra parentheses were necessary.

 

That argument must be > 0, so solve that inequality.

The solution to this inequality is (-infinity,-6) ∪ (5,infinity).

 

If x is in the interval [-6,5] then the argument (of log) is 0 or negative or undefined, which cannot be permitted.