# Simple math problem, exponents

• digipony
Thanks for the help! In summary, the student is trying to solve a physics problem and is unsure of their math. They ask for help with algebra and the person responds by suggesting that they use a rule that is - (v^a)^b = v^{ab}. The student then solves the problem using this information.

## 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!

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?

digipony said:
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 $$v^{^-\frac{1}{3}}=\frac{2P}{AN}$$ and you want to find $$v^{^+\frac{2}{3}}$$ ...?

note: the rule you want is - $(v^a)^b = v^{ab}$ so $$(v^{^-\frac{1}{3}})^2 = v^{(^-\frac{1}{3}\times \frac{2}{1})} = v^{^-\frac{2}{3}}$$... or:$$(v^{^-\frac{1}{3}})^{^-2} = v^{(^-\frac{1}{3}\times ^- \frac{2}{1})} = v^{^+\frac{2}{3}}$$

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Simon Bridge said:
Let me rewrite all that more clearly and you can tell me if I have understood you:

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

note: the rule you want is - $(v^a)^b = v^{ab}$ so $$(v^{^-\frac{1}{3}})^2 = v^{(^-\frac{1}{3}\times \frac{2}{1})} = v^{^-\frac{2}{3}}$$... or:$$(v^{^-\frac{1}{3}})^{^-2} = v^{(^-\frac{1}{3}\times ^- \frac{2}{1})} = v^{^+\frac{2}{3}}$$

Sorry for the format, I'm still trying to figure out how to use latex. Yes, that is right. I am starting out with $$v^{^-\frac{1}{3}}=\frac{2P}{AN}$$ and I wanted to use $(v^a)^b = v^{ab}$ to find out what $$v^{^\frac{2}{3}}$$ is equal to in terms of P,A and N.

Therefore, $$v^{^\frac{2}{3}}=(\frac{AN}{2P})^{^2}$$ correct?

digipony said:
Sorry for the format, I'm still trying to figure out how to use latex. Yes, that is right. I am starting out with $$v^{^-\frac{1}{3}}=\frac{2P}{AN}$$ and I wanted to use $(v^a)^b = v^{ab}$ to find out what $$v^{^\frac{2}{3}}$$ is equal to in terms of P,A and N.

digipony said:
Therefore, $$v^{^\frac{2}{3}}=(\frac{AN}{2P})^{^2}$$ correct?

Yes.
v-1/3 = (2P)/(AN)
=> v1/3 = (AN)/(2P)

Now you can square both sides, getting the equation you show, above.

Mark44 said:
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!

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?