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