Finding an eksponential function given 2 points

[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?

 

[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 :)

 

[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?

 

[HallsofIvy]

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.

Using your figures,
13.3(1.0055)^270= 58.5 instead of 59
13.3(1.0055)^420= 133.1 instead of 135.

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.

 

[Dr-NiKoN]

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

Thanks :)

 

[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?

 

[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:
$$c=\frac{y_{1}}{x_{1}^{r}}$$

 

[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?

 

[arildno]

It's called a power function.

 

[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:

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

 

[arildno]

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

what your book writes, is correct.

 

[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.

 

[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.

 

[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)

 

[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.

 

[arildno]

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

 

[Dr-NiKoN]

I highly doubt it :)

thanks a lot!