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## Main Question or Discussion Point

Does anyone know how you would go about solving:

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

- Thread starter natski
- Start date

- #1

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Does anyone know how you would go about solving:

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

- #2

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maybe this information would be helpful to me, and to others also who have more expertise.

- #3

uart

Science Advisor

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Sure. It's real easy.Does anyone know how you would go about solving:

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

If that's not the answer you were expecting then how about specifying which variable you want to solve for.

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- #5

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i guess natski already stated that he wants to solve for x.

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

- #6

uart

Science Advisor

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Whoops, that will teach me to read all of the post and not skip the last line.solving for x?

Sorry, that'ss a transcendental equation in x, so there will be no closed form solution.

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- #7

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ej uart how do u determine wheather it is transcendental equation in x?

- #8

uart

Science Advisor

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Oh yeah it's not transcendental in x. If y is an integer it's a polynomial in x.ej uart how do u determine wheather it is transcendental equation in x?

- #9

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- #10

uart

Science Advisor

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Well there might be some partiucular values of "y" for which it can work, but in general I don't think it's going to be possible.So I was hoping for a neater algebraic solution.

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