# Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

• May 2nd 2013, 11:40 AM
BreazaMathsB
Cooling the silly coffee !!!!! emmidiate help Due in a few Hours
Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

test y=ax^n and y=Ae^kx
1). i need to investigate a power model in the form y=ax^n by plotting my raw data and log x vs log y
2). i need to investigate a exponential modle in the form y=Ae^kx by plotting the raw data and x vs log y

i now need to use graphs and correlation coefficients to determine the equation of the model that fits your Data.?????

how do i find the equation... my text book only gose briefly and i missed the class due sports, i asked my teachers and friends but they only confused me more.
thank you for the fast help
• May 2nd 2013, 11:46 AM
HallsofIvy
Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours
What is your "raw data"? Presummably, each data point is (x, y). Find the two logarithms, and plot point (log(x), log(y)). Do those points seem to be close to a staright line?
The point is that if y= ax^n then lpg(y)= log(ax^n)= nlog(x)+ log(a) where log(y) is a linear function of log(x).

Similarly, if y= ae^(kx) the log(y)= kx+ log(a) where log(y) is a linear function of x.
• May 2nd 2013, 11:55 AM
BreazaMathsB
Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours
Hi thanks for being fast,
yes the raw data is in x and y x being time and y the difference in temperature as apart of a table we were given.
what i don't under stand is how do i find A and n in the fist and A and k in the next??
• May 2nd 2013, 11:57 AM
BreazaMathsB
Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours
and every time i graph with my graphics cal it doesn't pass the x axis
• May 2nd 2013, 02:10 PM
HallsofIvy
Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours
Is there any reason why "not passing the x-axis" is a problem? If you are referring to finding the x-intercept, you can extend the line until it does cross.

As I said before, the whole point is to try to get, in the first case, log(y)= A log(x)+ B, and, in the second case, log(y)= Ax+ B.

"A", in each case, is the slope of the line: choose two points, (log(x1), log(y1) and (log(x2), log(y2)), to find the slope $\displaystyle \frac{log(y2)- log(y1)}{log(x2)- log(x1)}$ and (x1, log(y1)), (x2, log(y2)) to find the slope $\displaystyle \frac{log(y2)- log(y1)}{x2- x1}$. You can find "B" by using that A with one point: B= log(y)- Alog(x) or B= log(y)- Ax.
• May 3rd 2013, 01:25 PM
BreazaMathsB
Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours
Thank you for that :)