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I have two points:

$$A(X_0,Y_0) $$ $$ B(X_1,Y_1)$$

And I need to find the function that creates an exponential growth between the two point. The fuction for exponential growth is $y = ab^x$ and then

$$y_0=a\cdot b^{x_0}$$ $$y_1=a\cdot b^{x_1}$$

and solve for $a$

$$a= \frac{b^{x_0}}{y_0}$$ $$a= \frac{b^{x_1}}{y_1}$$

because we want the same curve for the two equations we can say that they are equivalent:

$$\frac{b^{x_0}}{y_0} = \frac{b^{x_1}}{y_1}$$

or

$$y_1 \cdot b^{x_0} = y_0 \cdot b^{x_1}$$

and then

$$b^{x_1-x_0} = \frac{y_1}{y_0}$$ $$y_1 - b^{x_1-x_0} \cdot y_0= 0$$

Now let's plug in our a equation our solution:

$$y_1 - b^{x_1-x_0} \cdot y_0 = \frac{b^{x_0}}{y_0}$$

or

$$y_0\cdot y_1-b^{x_1-x_0}-b^{x_0} = 0$$

Now I'm stuck here because I have a point $C$ with $y_0 < Y_c > y_1$ and I need to know the $X_c$ that falls on the curve we have calculated.

Kahel
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1 Answers1

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Just divide the equation $y_0=ab^{x_0}$ by the equation $y_1=ab^{x_1}$ to find $b$. Then plug in the coordinates of either point to compute $a$.

Michael Hoppe
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  • You solution is brilliant. Can you show the process that you used to find this solution? – Kahel Oct 12 '13 at 16:28
  • It's not brilliant, but straightaway. Maybe you show where you have failed from once you've computed $b$ from $y_0/y_1=b^{x_0-x_1}$? – Michael Hoppe Oct 12 '13 at 16:43
  • Uhm I don't think I fail but I was seeking too much abstaction. I wanted something that cold be use for a exponential curve between two point A and B. If I'm not mistaken for a generic point (X,Y) on the exponential curve between (X0,Y0) and (X1,Y1) the function must be: Y = [logx0-x1(y0-y1)^x0/y0] * logx0-x1(y0-y1)^X – Kahel Oct 12 '13 at 17:31
  • So X will be X = LOG Base{[logx0-x1(y0-y1)^x0/y0] * logx0-x1(y0-y1)} of Y – Kahel Oct 12 '13 at 17:46