Quick Question: Write \frac{2\sqrt x+3}{x} in the Form 2x^p + 3x^q

[_Mayday_]

[SOLVED] Quick Question

Hey!

This question may tex ability :shy:

The Question
Write \frac{2\sqrt x+3}{x} in the form 2x^p + 3x^q where p and q are constants.

Attempt

\frac{2\sqrt x+3}{x} \times x

2x + 3x

I basically multiplied the initial equation by x to get rid of the denominator, and now I have what looks like a close to correct answer, but without p and q, either p or q could be one but I think the fact that they have asked for p and q would suggest that they powers must be different.

Thanks to any helpers =]

_Mayday_

 

[Avodyne]

Can you write 2\sqrt x/x in the form 2x^p?

Can you write 3/x in the form 3x^q?

Hint: don't multiply by x, this gives something different than what you started with (and are supposed to evaluate).

 

[Rashad9607]

Heed Avodyne's advice! Say you were asked to reduce 64/8 and you decided to multiply by 8 to get rid of the denominator. You now have a different number. Don't confuse this with equations, where you can perform an operation to both sides (like division) and maintain the equality.

 

[_Mayday_]

Can you write 2\sqrt x/x in the form 2x^p?

Can you write 3/x in the form 3x^q?

Hint: don't multiply by x, this gives something different than what you started with (and are supposed to evaluate).

x^{\frac{1}{2}}/x

3x^{-1}

I don't think I got the first bit correct...

Heed Avodyne's advice! Say you were asked to reduce 64/8 and you decided to multiply by 8 to get rid of the denominator. You now have a different number. Don't confuse this with equations, where you can perform an operation to both sides (like division) and maintain the equality.

Thanks

 

[CompuChip]

Basic algebra:
<br /> \frac{2\sqrt x+3}{x} = \frac{2\sqrt{x}}{x} + \frac{3}{x} = 2 \frac{\sqrt{x}}{x} + 3 \frac{1}{x}<br />

 

[_Mayday_]

I have the 3 but now, but I am still struggling with the root x. Please try and explain how it works.

Thank you very much for your help so far though, my maths is actually quite good but I have these little cracks I need filling in...

Thanks again.

 

[HallsofIvy]

You are correct that \sqrt{3}= 3^{1/2}. Now what about the 1/x? What do the "laws of exponents" tell you about x^m/x^n?

 

[_Mayday_]

Okay, I will take your post above as an example HallsofIvy? I tried looking for the root 3 but couldn't find it in my working!

I will take it that my answer is right:

x^\frac{1}{2}/x

which gives me...

x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}



Thanks Halls, I hope I am right...

 

[Snazzy]

Okay, I will take your post above as an example HallsofIvy? I tried looking for the root 3 but couldn't find it in my working!

I will take it that my answer is right:

x^\frac{1}{2}/x

which gives me...

x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}



Thanks Halls, I hope I am right...

Eh, your answer is right but oddly derived.

\frac{x^\frac{1}{2}}{x}=x^{\frac{1}{2}-1}=x^{\frac{-1}{2}}

 

[HallsofIvy]

It was just a typo. He meant
x^{\frac{1}{2}}\times x^{-1}= x^{-\frac{1}{2}}

 

[_Mayday_]

I appreciate all of your help everyone, thankyou!

_Mayday_