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#### RidiculousName

##### New member

- Jun 27, 2018

- 28

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

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

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- Thread starter
- #1

- Jun 27, 2018

- 28

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

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

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- Jan 30, 2018

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- Jun 27, 2018

- 28

\(\displaystyle x^2-5x>0 \) becomes \(\displaystyle x(x-5)>0\) or \(\displaystyle x^2>5x\) depending on what I do. I'm just not sure where to take it after that.Okay, we require:

\(\displaystyle x^2-5x>0\)

Can you factor the expression on the LHS?

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Observing that \(x^2-5x=x(x-5)\) tells us that the roots are:\(\displaystyle x^2-5x>0 \) becomes \(\displaystyle x(x-5)>0\) or \(\displaystyle x^2>5x\) depending on what I do. I'm just not sure where to take it after that.

\(\displaystyle x\in\{0,5\}\)

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?

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- Jun 27, 2018

- 28

So, since the coefficient of the squared root is positive, I can tell it's \(\displaystyle (\infty,0)\cup(5,\infty)\)?Observing that \(x^2-5x=x(x-5)\) tells us that the roots are:

\(\displaystyle x\in\{0,5\}\)

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?

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

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

and yes, this is correct.