Ok, I know how to do this but I don't have a graphing calculator now so. Question is:

(2, 13.2) (27, 40.5) (58, 82.7) (82, 172), find exponential function f(x) = ba^x

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- Sep 6th 2009, 06:51 PMSkyHighExponential modelling
Ok, I know how to do this but I don't have a graphing calculator now so. Question is:

(2, 13.2) (27, 40.5) (58, 82.7) (82, 172), find exponential function f(x) = ba^x - Sep 6th 2009, 07:07 PMProve It
Choose two points to substitute into the equation.

Then you'll have two equations in two unknowns to solve simultaneously.

In your case...

$\displaystyle 13.2 = ba^2$

$\displaystyle 40.5 = ba^{27}$

Notice that $\displaystyle ba^{27} = ba^2a^{25} = 13.2a^{25}$

So $\displaystyle 40.5 = 13.2a^{25}$

$\displaystyle \frac{40.5}{13.2} = a^{25}$

$\displaystyle \left(\frac{40.5}{13.2}\right)^{\frac{1}{25}} = a$.

I'm sure you can solve for b. - Sep 6th 2009, 07:15 PMmr fantastic
Note that $\displaystyle \log y = \log (b a^x) = \log b + x \log a$.

So my advice is to take plot the points ($\displaystyle \log y, ~ x)$ and draw the line of best fit through them. The gradient of this line can be used to get your best estimate of $\displaystyle a$ and the $\displaystyle \log y$-intercept can be used to get your best estimate of $\displaystyle b$. - Sep 6th 2009, 07:44 PMluobo
- Sep 7th 2009, 09:23 AMSkyHigh
Prove it, your method doesn't account for the last two points so it's not right I tried it.

This should be a very easy problem if I plot the points in the graphing calculator and find Exp Reg. Please someone with a graphing calculator kindly do it for me I don't have one. - Sep 7th 2009, 01:16 PMmr fantastic
- Sep 7th 2009, 03:46 PMSkyHigh
Thanks I don't know there's a thing a called Regression Applet before. I got my answer using it (Exponential Regression Applet).

- Sep 7th 2009, 04:29 PMProve It