# Exponential modelling

• Sep 6th 2009, 06:51 PM
SkyHigh
Exponential 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 PM
Prove It
Quote:

Originally Posted by SkyHigh
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

Choose two points to substitute into the equation.

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

$13.2 = ba^2$

$40.5 = ba^{27}$

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

So $40.5 = 13.2a^{25}$

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

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

I'm sure you can solve for b.
• Sep 6th 2009, 07:15 PM
mr fantastic
Quote:

Originally Posted by SkyHigh
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

Note that $\log y = \log (b a^x) = \log b + x \log a$.

So my advice is to take plot the points ( $\log y, ~ x)$ and draw the line of best fit through them. The gradient of this line can be used to get your best estimate of $a$ and the $\log y$-intercept can be used to get your best estimate of $b$.
• Sep 6th 2009, 07:44 PM
luobo
Quote:

Originally Posted by SkyHigh
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

4 points, 2 parameters. Are you doing a curve fitting?
• Sep 7th 2009, 09:23 AM
SkyHigh
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 PM
mr fantastic
Quote:

Originally Posted by SkyHigh
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.

I outlined how to do it by hand in my earlier post. If you've been given a question like this it's expected you have access to the necessary technology. If you do a Google search I'm sure you'll find an on-line regression applet suitable for your purposes.
• Sep 7th 2009, 03:46 PM
SkyHigh
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 PM
Prove It
Quote:

Originally Posted by SkyHigh
Thanks I don't know there's a thing a called Regression Applet before. I got my answer using it (Exponential Regression Applet).

I assumed the points lay on the curve, I didn't realise you were using a curve of best fit...