MHB Finding Domain for Natural Log with Exponent f(x)=ln(x^2−5x)

[RidiculousName]

I just asked a similar question, but I got help for that one, and now I am stumped again.

I need to find the domain for $$f(x) = ln(x^2-5x)$$

What's confusing me is how to deal with the exponent. I can't think of a way to get around it.

 

[MarkFL]

Okay, we require:

$$x^2-5x>0$$

Can you factor the expression on the LHS?

 

[HOI]

A product of two numbers is positive if and only if both factors have the same sign- both negative or both negative.

 

[RidiculousName]

Okay, we require:

$$x^2-5x>0$$

Can you factor the expression on the LHS?

$$ x^2-5x>0 $$ becomes $$x(x-5)>0$$ or $$x^2>5x$$ depending on what I do. I'm just not sure where to take it after that.

 

[MarkFL]

$$ x^2-5x>0 $$ becomes $$x(x-5)>0$$ or $$x^2>5x$$ depending on what I do. I'm just not sure where to take it after that.

Observing that \(x^2-5x=x(x-5)\) tells us that the roots are:

$$x\in\{0,5\}$$

Rather than testing intervals though, let's use what we know about the parabolic graphs of quadratic functions. We see the coefficient of the squared term is positive, which means the parabola opens upwards, and so, given that it has two real roots, we should expect the expression to be positive on either side of the two roots, and negative in between. Can you proceed?

 

[RidiculousName]

Observing that \(x^2-5x=x(x-5)\) tells us that the roots are:

$$x\in\{0,5\}$$

Rather than testing intervals though, let's use what we know about the parabolic graphs of quadratic functions. We see the coefficient of the squared term is positive, which means the parabola opens upwards, and so, given that it has two real roots, we should expect the expression to be positive on either side of the two roots, and negative in between. Can you proceed?

So, since the coefficient of the squared root is positive, I can tell it's $$(\infty,0)\cup(5,\infty)$$?

 

[MarkFL]

So, since the coefficient of the squared root is positive, I can tell it's $$(\infty,0)\cup(5,\infty)$$?

I believe you mean:

$$(-\infty,0)\cup(5,\infty)$$

and yes, this is correct. (Yes)