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What is the equation for plotting points on an exponential curve with fixed end points?

For example, if I want to plot 10 point along a curve that starts with 10,000 (x=1, y=10000) and ends with 30,000 (x=10, y=30000) the formula is y=10000*(1.129831^(x-1)).

But if I change 1.129831 to some different value, 2 for example, then the curve tops out at 5,120,000. I'd like to be able to change that base number to adjust the shape of the curve and still have the curve start at a y value of 10,000 and end at a y value of 30,000.

What I need to know is the formula for calculating points along a curve of varying shape, but fixed end points.

Thank you for your help!

2 Answers 2

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I think I phrased my question poorly.

I want to plot $y$ at any point $x$ on a curve across any given number of points $n$ greater than 0 using growth $b$ and I want to adjust the scale and base of the curve by specifying the minimum $y_1$ and maximum $y_n$.

I believe this is how to write the formula:

$y=y_1+\frac{(y_1b^{x-1}-y_1)(y_n-y_1)}{y_1b^{n-1}-y_1}$

In Excel you would use the following formula. I'm using named ranges in the formula to make it easier to follow.

=Minimum+(((Minimum*Growth^(Position-1))-Minimum)*(Maximum-Minimum)/((Minimum*Growth^(Periods-1))-Minimum))

Here's an example result set that describes a steep curve:

$n = 10$
$b = 2$
$(x_1, y_1) = (1, 10000)$
$(x_n, y_n) = (n, 30000)$

Gives:

$(x_1, y_1) = (1, 10000)$
$(x_2, y_2) = (2, 10039)$
$(x_3, y_3) = (3, 10117)$
$(x_4, y_4) = (4, 10274)$
$(x_5, y_5) = (5, 10587)$
$(x_6, y_6) = (6, 11213)$
$(x_7, y_7) = (7, 12466)$
$(x_8, y_8) = (8, 14971)$
$(x_9, y_9) = (9, 19980)$
$(x_{10}, y_{10}) = (10, 30000)$

Here's another example result set that describes a shallow curve:

$n = 10$
$b = 1.1$
$(x_1, y_1) = (1, 10000)$
$(x_n, y_n) = (n, 30000)$

Gives:

$(x_1, y_1) = (1, 10000)$
$(x_2, y_2) = (2, 11473)$
$(x_3, y_3) = (3, 13093)$
$(x_4, y_4) = (4, 14875)$
$(x_5, y_5) = (5, 16835)$
$(x_6, y_6) = (6, 18992)$
$(x_7, y_7) = (7, 21364)$
$(x_8, y_8) = (8, 23973)$
$(x_9, y_9) = (9, 26843)$
$(x_{10}, y_{10}) = (10, 30000)$

Hopefully someone else will find this useful. Maybe someone can even find a way to simplify it.

Thanks!

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    Got it. I changed the title and tag of the question. I figured I hadn't phrased the question well, and I appreciate your help.2011-04-03
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If you have two points, you have two equations in two unknowns. In the case you cite, you have the points $(x_1,y_1)=(1,10000)$ and $(x_2,y_2)=(10,30000).$ If you claim the equation is $y=ka^x$ by taking the log you can solve for $k$ and $a$ and you don't have any choice. (note that you can absorb the $-1$ in the exponent in your solution into $k$-that just shifts $x_1$ down to $0$)

If you know the $y$ values of interest and want to specify $a$ (as $2$ in your example) you need to provide another degree of freedom. Maybe letting $x_2$ vary will satisfy your need. In that case, you need $a^{(x_2-x_1)}=\frac{y_2}{y_1}$, so in your example if $\frac{y_2}{y_1}=3$ and $a=2, x_2-x_1=\log_23$, so you can space $10$ points from $1$ to $1+\log_23$

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    Thank you for your answer Ross. I think I phrased my question poorly and figured out how to get what I was after. I also attempted to write out the equation and am giving the Excel equivalent so that others can benefit from it.2011-04-03