# How to solve V=B+CT(V^N) for V?

• thiago.omena
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.

#### thiago.omena

Does anyone know how to solve V=B+CT(V^N) for V?

I don't understand how to get there.

Hey thiago.omena and welcome to the forums.

Are these matrices or just plain old numbers?

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.

thiago.omena said:
Does anyone know how to solve V=B+CT(V^N) for V?
I don't understand how to get there.
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.

Btw, it's better to use the formatting this site provides. If you click Go Advanced you'll see a toolbar with e.g. X2 for superscript. Select the text to be superscripted and click the icon:
V=B+CT(VN)

$V= B+ CTV^N$ is the same as the polynomial equation $CTV^N- V+ B= 0$. There is NO general formula for solving a polynomial equation of degree greater than 5. In fact, it was proved about a hundred years ago that there cannot be a formula (involving only roots of numbers) for such equations.

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

thiago.omena said:
Does anyone know how to solve V=B+CT(V^N) for V?

I don't understand how to get there.

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.

## 1. How do I solve for V in the equation V=B+CT(V^N)?

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).

## 2. What does the V^N term represent in this equation?

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.

## 3. Is there a specific order in which I should solve for V?

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.

## 4. Can I solve for V if I don't know the values of B, C, or N?

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.

## 5. How can I check if I have solved for V correctly?

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.