1. ## 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

2. 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.

3. 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 $b$.

4. 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?

5. 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.

6. 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.

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

8. 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...