Find D: [(A - D)^P] / [(B - D)^R] = (C - D)^(P - R)

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Discussion Overview

The discussion revolves around the challenge of isolating the variable D in the equation [(A - D)^P] / [(B - D)^R] = (C - D)^(P - R). Participants explore various methods and related concepts, including logarithmic manipulation and the application of Pascal's Triangle in solving for unknown exponents.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about isolating D in the given equation.
  • Another suggests taking logarithms to potentially simplify the equation.
  • A participant introduces a complex definition related to Pascal's Triangle, aiming to create a general method for solving unknown exponents.
  • There is a question about what the unknown exponents refer to, specifically P and R.
  • One participant mentions a different equation related to fitting data points, which diverges from the original equation discussed.
  • Another participant provides a simplification of their approach to fitting a curve to data points, leading to a new equation that relates different variables.
  • A request is made for clearer formatting of the mathematical expressions to enhance readability.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to isolate D, and there are multiple competing views regarding the relevance of different equations and approaches to the problem.

Contextual Notes

The discussion includes various assumptions and dependencies on definitions, particularly regarding the unknown exponents and the equations being referenced. Some steps in the mathematical reasoning remain unresolved.

Raisintoe
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How do I get D by itself? This one's got me baffled

[(A - D)^P] / [(B - D)^R] = (C - D)^(P - R)
 
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Raisintoe said:
How do I get D by itself? This one's got me baffled

[(A - D)^P] / [(B - D)^R] = (C - D)^(P - R)
Take a lotta logs and see if anything shakes out.
 
SteamKing said:
Take a lotta logs and see if anything shakes out.
Ha ha ha! I've been to the bathroom enough times while trying to figure this out.
 
I have been trying to come up with a definition for Pascal's Triangle so that I can create a general way to solve for unknown exponents. All that I've been able to come up with so far is (1 - N + (N^2 - N)/2 - [1/2∑(n=2 to N) N(N - 2n + 1) + n(n - 1)] . . . ) for Pascal's Triangle of Coeficients
 
Raisintoe said:
I have been trying to come up with a definition for Pascal's Triangle so that I can create a general way to solve for unknown exponents. All that I've been able to come up with so far is (1 - N + (N^2 - N)/2 - [1/2∑(n=2 to N) N(N - 2n + 1) + n(n - 1)] . . . ) for Pascal's Triangle of Coeficients
Unknown exponents of what?
 
SteamKing said:
Unknown exponents of what?
My exponents, P and R
 
Basically you have x^r * y^(p-r) = 1. Without knowing anything about r and p it'll going to be hard. Are you dealing with economic indexes?
 
fresh_42 said:
Basically you have x^r * y^(p-r) = 1. Without knowing anything about r and p it'll going to be hard. Are you dealing with economic indexes?
I don't know what economic indexes are, but I am trying to solve for two unknowns in the common equation: V(t) = Vf + (Vi - Vf)*e^(-t/T) where Vf and T are unknown. I am trying to fit this curve to data points that I have collected.
 
Raisintoe said:
I don't know what economic indexes are, but I am trying to solve for two unknowns in the common equation: V(t) = Vf + (Vi - Vf)*e^(-t/T) where Vf and T are unknown. I am trying to fit this curve to data points that I have collected.
That's a completely different equation than what you had in the OP.
 
  • #10
SteamKing said:
That's a completely different equation than what you had in the OP.

I want to fit the curve to three points, one point gives my Vi, the other two are my different V(t)s. I simplified to get: T = -t/[ln((V(t) - Vf)/(Hi - Hf))]. Now I can set two equations equal to each other: -ta/[ln((V(ta) - Vf)/(Hi - Hf))] = -tb/[ln((V(tb) - Vf)/(Hi - Hf))].
This simplified to [(H(ta) - Hf)/(Hi - Hf)]^tb = [(H(tb) - Hf)/(Hi - Hf)]^ta. Then [(H(ta) - Hf)^tb]/[(H(tb) - Hf)^ta] = (Hi - Hf)^(tb-ta)
 
  • #11
Raisintoe said:
I have been trying to come up with a definition for Pascal's Triangle so that I can create a general way to solve for unknown exponents. All that I've been able to come up with so far is (1 - N + (N^2 - N)/2 - [1/2∑(n=2 to N) N(N - 2n + 1) + n(n - 1)] . . . ) for Pascal's Triangle of Coeficients
It would help if you could Tex this, making it easier to read.
 

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