Solving a polynomial of arbitrary powers

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In summary, the equation W*C^(Z/A) = X*C^(Y/A) - D*X + D*W can be solved for A in terms of the other constants by substituting k=C^(1/A) and rearranging the equation. However, there is no general formula for solving polynomials, so the solution may not always be possible.
  • #1
davee123
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W*C^(Z/A) = X*C^(Y/A) - D*X + D*W

How do you go about solving for A?
Known: C>1
Also, if it makes it easier (it shouldn't) you can assume that D=1.

If Y and Z are known, I figure you can substitute k=C^(1/A), which turns it into a more trivial polynomial. But if Y and Z *aren't* known, I'm at a loss.

For the sake of reference, here's where the problem comes from:

I want a logarithmic function that I can use in a program I'm writing. Therefore, I'm guessing that I want an equation of the form:

f(x) = A*log(B*x + D) (that's log base C)

And I know that:
f(0) = 0
f(W) = Y
f(X) = Z

Since F(0) = 0, I can assume that D = 1. No problem.
And I don't really care what C is, nor do I think I ought to-- if it matters, I'm just assuming that C = e.

Plugging in W and solving for B in terms of A gives me:

B = (C^(Y/A) - D)/W

Plugging in X and solving for A gives me something I can't solve-- at least not in the general case:

W*C^(Z/A) = X*C^(Y/A) - D*X + D*W

[edit]
And if I do the reverse, and try solving for B instead of A, I get:

(B*W+D)^Z = (B*X+D)^Y

Which I similarly can't figure out how to solve for B.
[/edit]

DaveE
 
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  • #2
First, that's not a polynomial. Second, since it has nothing to do with "Linear and Abstract Algebra", I am moving this to "General Math".

If Y and Z are known, I figure you can substitute k=C^(1/A), which turns it into a more trivial polynomial. But if Y and Z *aren't* known, I'm at a loss.
I don't know why you would say that. You can solve FOR A in terms of the other constants. If you let k= C1/A you get
WkZ= XkY- DX+ DY

Now, if Z and Y are positive integers, that IS a polynomial. However, there are no general formulas for solving polynomials.
 
  • #3
HallsofIvy said:
First, that's not a polynomial. Second, since it has nothing to do with "Linear and Abstract Algebra", I am moving this to "General Math".

D'oh! Yeah, I wasn't sure-- I figured I'd look for an "Algebra" forum, but that didn't exist.

HallsofIvy said:
I don't know why you would say that. You can solve FOR A in terms of the other constants. If you let k= C1/A you get
WkZ= XkY- DX+ DY

Now, if Z and Y are positive integers, that IS a polynomial.

Ahh, ok. I never realized polynomials required positive integers. Hm. I suppose as long as Z and Y are rational (and both positive or both negative), you could choose a better substitution that still turns it into a polynomial, but of course, there's still no guarantee.

DaveE
 

What is a polynomial of arbitrary powers?

A polynomial of arbitrary powers is an algebraic expression that contains variables raised to any power, such as x3, x5, or xn. These powers can be positive or negative integers, or even fractions.

How do you solve a polynomial of arbitrary powers?

To solve a polynomial of arbitrary powers, you can use various techniques such as factoring, completing the square, or the quadratic formula. You can also use synthetic division or the rational root theorem for polynomials with rational coefficients.

What are the steps to solve a polynomial of arbitrary powers?

The general steps to solve a polynomial of arbitrary powers are:

  1. Arrange the terms in descending order of powers.
  2. Factor out the greatest common factor (GCF), if possible.
  3. Use appropriate factoring techniques to factor the polynomial, if possible.
  4. Apply the zero product property to find the roots or solutions of the polynomial.
  5. Check your solutions by plugging them back into the original polynomial.

What are some common mistakes when solving a polynomial of arbitrary powers?

Some common mistakes when solving a polynomial of arbitrary powers include:

  • Forgetting to apply the order of operations.
  • Not factoring out the GCF.
  • Incorrectly applying factoring techniques.
  • Making errors while using the quadratic formula or completing the square.
  • Forgetting to check solutions at the end.

What are some real-life applications of solving polynomials of arbitrary powers?

Solving polynomials of arbitrary powers is used in various fields such as engineering, physics, computer science, and economics. Some real-life applications include:

  • Calculating trajectories of projectiles in physics.
  • Designing computer algorithms for image processing.
  • Modeling economic growth and population growth.
  • Designing complex structures in engineering.

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