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I have two intervals: $X=[A,B]$ and $X'=[C,D]$.

If I'd like to map $X$ to $X'$, I usually use this equation:

$$f(t)= \frac{D-C} {B-A} t + \frac{(BC - AD)} {B-A}$$

where $t$ is the time.

However this is linear mapping. I would like to do the mapping exponentially. Could someone help me in this?

I appreciate you help.

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I just reformatted your equations. You might want to check them over to make sure I didn't screw them up somehow. :) –  Micah Mar 28 '13 at 0:18

1 Answer 1

I assume by "exponentially" you mean you want a map of the form $f(t)=ke^{rt}$. So we get the equations $$C=ke^{Ar},\qquad D=ke^{Br}$$ Then $$C/D=e^{(A-B)r}$$ so $$r={\log C-\log D\over A-B}\tag1$$ and $k=Ce^{-kr}$ where $r$ is given by (1).

EDIT: As Ross points out in the comments, that was meant to be $k=Ce^{-Ar}$.

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Thanks for your reply. So how do I calculate f(t) from A B C and D? –  user951793 Mar 27 '13 at 12:53
    
How is k = Ce (pow) -kr –  user951793 Mar 27 '13 at 13:57
1  
@user951793: I believe $k=Ce^{-Ar}$ Once you have $k,r$ you have everything you need for $f(t)$ –  Ross Millikan Mar 28 '13 at 0:49
    
@RossMillikan I tried the equation with calculating k value as you suggested, but I get the following over the time: Scale:Infinity Scale:Infinity Scale:Infinity My test case is: A = 100; B = 3100; C = 0; D = 400; –  user951793 Mar 28 '13 at 8:13
    
@user951793: that is because you are trying to take the log of zero with $C=0$. The way $f(t)$ is defined, it cannot be zero. –  Ross Millikan Mar 28 '13 at 13:04

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