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

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To isolate D in the equation [(A - D)^P] / [(B - D)^R] = (C - D)^(P - R), logarithmic manipulation is suggested, although the complexity of the exponents P and R complicates the process. The discussion shifts to a different equation, V(t) = Vf + (Vi - Vf)e^(-t/T), where Vf and T are unknowns, and the user seeks to fit this curve to collected data points. A simplification leads to a new equation relating the variables, but it diverges from the original problem. The conversation highlights the challenges of solving for unknowns without additional context or definitions. Overall, the thread emphasizes the intricacies of mathematical equations involving unknown variables and the need for clarity in the problem statement.
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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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