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Quick Question

  • Thread starter _Mayday_
  • Start date
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[SOLVED] Quick Question

Hey!

This question may tex ability :shy:

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

Attempt :eek:

[tex]\frac{2\sqrt x+3}{x} \times x[/tex]

[tex]2x + 3x[/tex]

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_
 

Answers and Replies

Avodyne
Science Advisor
1,396
85
Can you write [itex]2\sqrt x/x[/itex] in the form [itex]2x^p[/itex]?

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

Hint: don't multiply by x, this gives something different than what you started with (and are supposed to evaluate).
 
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.
 
795
0
Can you write [itex]2\sqrt x/x[/itex] in the form [itex]2x^p[/itex]?

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

Hint: don't multiply by x, this gives something different than what you started with (and are supposed to evaluate).
[tex]x^{\frac{1}{2}}/x[/tex]

[tex]3x^{-1}[/tex]

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
Science Advisor
Homework Helper
4,284
47
Basic algebra:
[tex]
\frac{2\sqrt x+3}{x} = \frac{2\sqrt{x}}{x} + \frac{3}{x} = 2 \frac{\sqrt{x}}{x} + 3 \frac{1}{x}
[/tex]
 
795
0
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
Science Advisor
Homework Helper
41,732
893
You are correct that [itex]\sqrt{3}= 3^{1/2}[/itex]. Now what about the 1/x? What do the "laws of exponents" tell you about [itex]x^m/x^n[/itex]?
 
795
0
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! :rofl:

I will take it that my answer is right:

[tex]x^\frac{1}{2}/x[/tex]

which gives me...

[tex]x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}[/tex]



Thanks Halls, I hope I am right...
 
458
0
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! :rofl:

I will take it that my answer is right:

[tex]x^\frac{1}{2}/x[/tex]

which gives me...

[tex]x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}[/tex]



Thanks Halls, I hope I am right...
Eh, your answer is right but oddly derived.

[tex]\frac{x^\frac{1}{2}}{x}=x^{\frac{1}{2}-1}=x^{\frac{-1}{2}}[/tex]
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
It was just a typo. He meant
[tex]x^{\frac{1}{2}}\times x^{-1}= x^{-\frac{1}{2}}[/tex]
 
795
0
I appreciate all of your help everyone, thankyou!

_Mayday_
 

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