Question 246026
{{{f(x)=log(7, (2-(x+5)/(x-6)))}}}<br>
There are two issues to address here:<ul><li>Denominators may not be zero.</li><li>Arguments of logarithms, no matter what the base, must not be zero or negative.</li></ul>
Denominator.
Let's find what x value makes the denominator zero:
x - 6 = 0
Add 6 to both sides:
x = 6
This is a value we <b>cannot</b> allow for x. It must be excluded from the domain.<br>
Argument of the logarithm.
We want a positive argument:
{{{2-(x+5)/(x-6) > 0}}}
There are different ways to solve this. One way is to subtract the terms on the left. Of course we need common denominators first:
{{{2((x-6)/(x-6))-(x+5)/(x-6) > 0}}}
{{{(2x-12)/(x-6)-(x+5)/(x-6) > 0}}}
Subtract (Be careful when subtracting!) :
{{{(x -17)/(x-6) > 0}}}
This says we have a positive fraction. When do we get positive fractions? Answer: When the numerator and denominator are both positive or when they are both negative. This idea, which you probably knew well, can be expressed in the algebraic form:
({{{x-17 > 0}}} and {{{ x-6 > 0}}}) or ({{{x-17 < 0}}} and {{{ x-6 < 0}}})
(Look at this and see if you can understand how this says "both positive or both negative".)
Now we solve this compound inequality.
({{{x > 17}}} and {{{ x > 6}}}) or ({{{x < 17}}} and {{{ x < 6}}})
In the first pair, in order for x to be greater than 17 and 6, it would have to be greater than 17. In the last pair, in order for x to be less than 17 and 6, it would have to be less than 6. So
({{{x > 17}}} and {{{ x > 6}}}) or ({{{x < 17}}} and {{{ x < 6}}})
simplifies to
{{{x > 17}}} or {{{ x < 6}}}<br>
Since {{{x > 17}}} or {{{ x < 6}}} excludes the number which makes the denominator zero, this is our domain.