Attachment 23671Hello everyone!

I really appreciate anyone that can help me with this problem:

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- Apr 20th 2012, 08:28 PMketerWorking with Real DATA HELP!
Attachment 23671Hello everyone!

I really appreciate anyone that can help me with this problem: - Apr 20th 2012, 11:15 PMProve ItRe: Working with Real DATA HELP!
If it's real data, it's highly unlikely that the data will fit an exponential model exactly.

It might help if you realise

$\displaystyle \displaystyle \begin{align*} C\,a^x &= C\,e^{\ln{\left(a^x\right)}} \\ &= C\,e^{x\ln{a}} \\ &= C\left(e^x\right)^{\ln{a}} \end{align*}$

And now when you apply

$\displaystyle \displaystyle \begin{align*} e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \end{align*}$ to a desired level of accuracy, you should be able to fit a least squares polynomial. - Apr 21st 2012, 04:10 AMBobPRe: Working with Real DATA HELP!
Think of the required equation as y=Ca

^{x}where values of y are the CO_{2}levels and then take logs.

After 'simplifying' (using a couple of rules for logarithms), the equation becomes

log(y) = log(C) + xlog(a).

(It doesn't matter what base of logs that you use so long as you are consistent throughout.)

If log(y), (vertical axis) is plotted against x, what you should find (assuming that the law you are trying to fit is a reasonable one) is that your points lie roughly along a straight line.

Draw in what you think is the best straight line and then read off the intercept on the vertical axis (which will be the value of log(C)) and calculate the slope (which will be the value of log(a)).

Antilog to find the values of C and a and substitute into the original equation.

Finally, check to see that the equation gives (approximately) the correct values for the CO_{2}levels for the x values.