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Simple math problem, exponents

  • Thread starter digipony
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  • #1
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Homework Statement



I am working on a physics problem, and am unsure of my math at one point. The physics problem is irrelevant, as I just want to verify my math. I am in calc3 and diff eq., I am just a little fuzzy/unsure of my algebra sometimes. Could someone please tell me if the work I have below follows the proper math/algebra rules? Thanks!


Homework Equations



n/a

The Attempt at a Solution



Let's say I have v^-1/3 = 2P/(AN) and I want to find out what v^2/3 is equal to. In order to do this, can I just raise both sides of the equation to the -2 power so that I get this: v^-2/3 = (2P/(AN))^-2 which simplifies to this: v^-2/3 = (AN/(2P))^2 right?
 

Answers and Replies

  • #2
Simon Bridge
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Let's say I have v^-1/3 = 2P/(AN) and I want to find out what v^2/3 is equal to. In order to do this, can I just raise both sides of the equation to the -2 power so that I get this: v^-2/3 = (2P/(AN))^-2 which simplifies to this: v^-2/3 = (AN/(2P))^2 right?
Let me rewrite all that more clearly and you can tell me if I have understood you:

You have [tex]v^{^-\frac{1}{3}}=\frac{2P}{AN}[/tex] and you want to find [tex]v^{^+\frac{2}{3}}[/tex] ...?

note: the rule you want is - [itex](v^a)^b = v^{ab}[/itex] so [tex](v^{^-\frac{1}{3}})^2 = v^{(^-\frac{1}{3}\times \frac{2}{1})} = v^{^-\frac{2}{3}}[/tex]... or:[tex](v^{^-\frac{1}{3}})^{^-2} = v^{(^-\frac{1}{3}\times ^- \frac{2}{1})} = v^{^+\frac{2}{3}}[/tex]
 
Last edited:
  • #3
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Let me rewrite all that more clearly and you can tell me if I have understood you:

You have [tex]v^{^-\frac{1}{3}}=\frac{2P}{AN}[/tex] and you want to find [tex]v^{^+\frac{2}{3}}[/tex] ...?

note: the rule you want is - [itex](v^a)^b = v^{ab}[/itex] so [tex](v^{^-\frac{1}{3}})^2 = v^{(^-\frac{1}{3}\times \frac{2}{1})} = v^{^-\frac{2}{3}}[/tex]... or:[tex](v^{^-\frac{1}{3}})^{^-2} = v^{(^-\frac{1}{3}\times ^- \frac{2}{1})} = v^{^+\frac{2}{3}}[/tex]
Sorry for the format, I'm still trying to figure out how to use latex. Yes, that is right. I am starting out with [tex]v^{^-\frac{1}{3}}=\frac{2P}{AN}[/tex] and I wanted to use [itex](v^a)^b = v^{ab}[/itex] to find out what [tex]v^{^\frac{2}{3}}[/tex] is equal to in terms of P,A and N.
 
  • #4
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Therefore, [tex]v^{^\frac{2}{3}}=(\frac{AN}{2P})^{^2}[/tex] correct?
 
  • #5
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Sorry for the format, I'm still trying to figure out how to use latex. Yes, that is right. I am starting out with [tex]v^{^-\frac{1}{3}}=\frac{2P}{AN}[/tex] and I wanted to use [itex](v^a)^b = v^{ab}[/itex] to find out what [tex]v^{^\frac{2}{3}}[/tex] is equal to in terms of P,A and N.
Therefore, [tex]v^{^\frac{2}{3}}=(\frac{AN}{2P})^{^2}[/tex] correct?
Yes.
v-1/3 = (2P)/(AN)
=> v1/3 = (AN)/(2P)

Now you can square both sides, getting the equation you show, above.
 
  • #6
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Yes.
v-1/3 = (2P)/(AN)
=> v1/3 = (AN)/(2P)

Now you can square both sides, getting the equation you show, above.
Great! Thank you!
 
  • #7
Simon Bridge
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OR you can raise both sides to the power of (-2) and then invert the RHS - however, it is a useful discipline to invert first: less chance of confusion.

No worries eh?
 

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