# Finding an eksponential function given 2 points

1. Oct 31, 2004

### Dr-NiKoN

Given the two points: (270, 59) and (420, 135) I want to find an exponential function that goes trough these two points.
So I have:
$ca^{260} = 50$ and $ca^{420} = 135$

I then divide these two functions:

$\frac{ca^{270}}{ca^{420}} = \frac{59}{135}$

$a^{270-420} = \frac{59}{135}$

$a = (\frac{59}{135})^{\frac{1}{270-420}}$

Is this the correct way for finding a and c, thus giving me a eksponential function?

2. Oct 31, 2004

### SomeRandomGuy

Your method is correct, however, your ;ast two steps use 270-420 instead of 260-420. You probably just copied it wrong by accident, no big deal. By the way, it's exponential :)

3. Oct 31, 2004

### Dr-NiKoN

Ok, then I need help.
It's 270 btw.

$a = (\frac{59}{135})^{\frac{1}{270-420}}$

$a= 1.0055$

Now finding c:
$ca^{270} = 59$

$c = \frac{59}{a^{270}} = \frac{59}{1.0055^{270}} = 13.3$

But:
$f(x) = 13.3 * 1.0055^x$
Isn't correct.

I'm not sure where I'm going wrong here :(
Shouldn't a or c be less than 1?

Last edited: Oct 31, 2004
4. Nov 1, 2004

### HallsofIvy

Staff Emeritus
No, since this is an increasing function, a cannot be less than 1. I don't know why you think a or c must be less than 1.

It looks to me like you are just rounding off too much. I get
a= 1.005533502 and then c= 13.29795470.

Using those values, c a^270= 59.00000000 and c a^420= 135.00000000.

5. Nov 1, 2004

### Dr-NiKoN

Sorry guys, the confusion was due to me using an incorrect number.

Thanks :)

Last edited: Nov 1, 2004
6. Nov 1, 2004

### Dr-NiKoN

I'll use this thread instead of creating a new one. I'm working on creating emphiric functions given various data-sets.

Linear functions and as shown above exponential functions are fine.
Now I'm working with functions that are similar to exponential functions, but using log.
I want a function of the form: $f(x) = c * x^r$

The data set:
$(\log{x_0}, \log{y_0}), (\log{x_1}, \log{y_1}).$

There might obviously be more points. From these points, we try to draw out a straight line as possible on a graph. Then find the graphs slope(?) graphically with a ruler. ie
$\frac{\Delta x}{\Delta y} = r$

Now we have:
$c * x_0^r = y_0 \rightarrow c = \frac{y_0}{x_0^r}$

I've tried this out with various data, but my function is always very inaccurate. It's 100% for the point I use to find 'c', but for any other point the result might be as much as 50% off. I know it's an emphiric function, but I would expect I would be able to get it more accurate.

Is this normal?

7. Nov 1, 2004

### arildno

Whatever have you been doing??
We have:
$$y(x)=cx^{r}$$
or, EQUIVALENTLY:
$$log(y)=rlog(x)+log(c)$$
You are to interpolate, using initial conditions:
$$log(y_{0})=rlog(x_{0})+log(c)$$
$$log(y_{1})=rlog(x_{1})+log(c)$$
This yields:
$$r=\frac{log(\frac{y_{1}}{y_{0}})}{log(\frac{x_{1}}{x_{0}})}$$
$$log(c)=log(y_{1})-rlog(x_{1})=log(\frac{y_{1}}{x_{1}^{r}})$$
Or:
$$c=\frac{y_{1}}{x_{1}^{r}}$$

Last edited: Nov 1, 2004
8. Nov 1, 2004

### Dr-NiKoN

Ah, that works like a charm.

I'm guessing my error was that I was still treating it as an exponential function, just using log on the values for each point.

What is a function of the form f(x) = ax^r called by the way? Is this also a linear function?

9. Nov 1, 2004

### arildno

It's called a power function.

10. Nov 2, 2004

### Dr-NiKoN

I'm sorry if this is a stupid question, but my book doesn't mention it.

Is this correct?

$\frac{\log(x_0)}{\log(x_1)} = \log(\frac{x_1}{x_0})$

Because my book says:

$\log(\frac{x_0}{x_1}) = \log(x_0) - \log(x_1)$

Last edited: Nov 2, 2004
11. Nov 2, 2004

### arildno

No it is not!
I can't see that I have done that mistake.

what your book writes, is correct.

12. Nov 2, 2004

### Dr-NiKoN

I'm sure what you have written is correct, I'm just having trouble getting from:

$log(y_{1})=rlog(x_{1})+log(c)$
$log(y_{0})=rlog(x_{0})+log(c)$

to

$r=\frac{log(\frac{y_{1}}{y_{0}})}{log(\frac{x_{1}} {x_{0}})}$

Have you done:

$\log(y_0) - r\log(x_0) = \log(y_1) - r\log(x_1)$

To find r? I don't understand what you mean by interpolating, or what step you did.

13. Nov 2, 2004

### arildno

I have subtracted the left-hand SIDES from each other, and equated that to the subtraction of the right-hand sides from each other.

14. Nov 2, 2004

### arildno

Note "interpolation" means to fit given data to some type of curve (in this case, we determine what sort of exponential curve fits two given data points)

15. Nov 2, 2004

### Dr-NiKoN

$\log(y_0) - r\log(x_0) = \log(y_1) - r\log(x_1)$

$r\log(x_1) - r\log(x_0) = \log(y_1) - \log(y_0)$

$r(\log(x_1) - \log(x_0)) = \log(y_1) - \log(y_0)$

$r * \log(\frac{x_1}{x_0}) = \log(\frac{y_1}{y_0})$

$r = \frac{\log(\frac{y_1}{y_0})}{\log(\frac{x_1}{x_0})}$

Yay :)

I was just wondering if this was the same way you did it. It seems so elemental, and I thought you where using other elements of log to get there.

16. Nov 2, 2004

### arildno

No, your understanding of logs is clearly as "deep" as my own..

Last edited: Nov 2, 2004
17. Nov 2, 2004

### Dr-NiKoN

I highly doubt it :)

thanks a lot!