- #1
davee123
- 672
- 4
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
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
Last edited: