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The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))

I don't understand how to get there.

Please help me out. Thank you

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- Thread starter thiago.omena
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In summary, the conversation discusses how to solve the equation V=B+CT(V^N) for V, with some users providing a potential solution and others questioning its validity and suggesting alternative approaches. Ultimately, it is noted that there is no general formula for solving polynomial equations of degree greater than 5, making an exact solution for this equation unlikely.

- #1

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The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))

I don't understand how to get there.

Please help me out. Thank you

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- #2

Science Advisor

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Hey thiago.omena and welcome to the forums.

Are these matrices or just plain old numbers?

Are these matrices or just plain old numbers?

- #3

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chiro said:Hey thiago.omena and welcome to the forums.

Are these matrices or just plain old numbers?

In the question? They are just variables. I am trying to isolate the V.

- #4

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What makes you think that's the answer? It doesn't look right to me. In fact, I do not believe there is a solution in closed form.thiago.omena said:Does anyone know how to solve V=B+CT(V^N) for V?

The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))

I don't understand how to get there.

Btw, it's better to use the formatting this site provides. If you click Go Advanced you'll see a toolbar with e.g. X

V=B+CT(V

- #5

Science Advisor

Homework Helper

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And it is easy to show that your [itex]V=[B^{1-N}+C(N-1)T]^(1/(1-N))[/itex] is NOT a general solution to the equation. In the case that C= T= 1, B= -2 and N= 2, the equation becomes [itex]V= -2+ V^2[/itex] which is the same as [itex]V^2- V- 2= (V- 2)(V+ 1)= 0[/itex] which has roots 2 and -1. But your formula gives [itex]V= [(-2)^{-1}+ (1)(-1)(1)]^{1/-1}= [-1/2+ -1]^{-1}= [-3/2]^{-1}= -2/3[/itex] which is NOT a solution.

- #6

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thiago.omena said:

The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))

I don't understand how to get there.

Please help me out. Thank you

Is this a thermodynamic equation? If it is, you're likely looking for an asymptotic solution for V >> 1 and N >> 1. As has been said, solving the equation exactly is not possible in general, but an approximate solution in the large V and/or N limits may be possible, and is perhaps the answer you claim to have.

To solve for V in this equation, you can use algebraic manipulation to isolate V on one side of the equation. Begin by subtracting B from both sides to get V- B = CT(V^N). Then, divide both sides by CT to get (V-B)/CT = V^N. Finally, take the Nth root of both sides to get the final solution: V = (V-B)/CT^(1/N).

The V^N term represents V raised to the power of N. In other words, V is multiplied by itself N times. This can also be written in shorthand as V^n.

No, there is no specific order in which you need to solve for V. As long as you use algebraic manipulation correctly and follow the correct steps, you will arrive at the correct solution regardless of the order in which you solve for V.

No, in order to solve for V in this equation, you need to have values for B, C, and N. Without these values, you will not be able to isolate V and find a specific solution.

To check if you have solved for V correctly, you can substitute your solution back into the original equation and see if it holds true. If both sides of the equation are equal, then you have solved for V correctly.

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