SOLUTION: Hello! Here is the information I have been given, an explanation of what I need to do, what I understand and what I don't. Please help! Thanks for looking at this problem! Tim

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hello! Here is the information I have been given, an explanation of what I need to do, what I understand and what I don't. Please help! Thanks for looking at this problem! Tim      Log On


   



Question 64047: Hello!
Here is the information I have been given, an explanation of what I need to do, what I understand and what I don't.
Please help!
Thanks for looking at this problem!
Time /Temp
5 /55.7
15 /49.1
25 /42.8
35 /37
50 /29.9
60 /25.8
70 /22.4
80 /19.2
90 /16.4
100 /14.1
120 /10.3
140 /7.4
150 /6.1
170 /4
180 /3.3
The instructions tell me to calculate a linear regression line and then and from that to write an exponential equation that will match the original data.
I get how to figure out the linear regression line.
in y=a%2Bbx form I get y=47.4021-.2879x
I don't understand how to derive an exponential equation from this information.
Thanks Again

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
One way to get the exponential regression equation is to
use a calculator with that function programmed into it.
The TI-83 is such a calculator.
Using it I get this exponential equation:
y=(65.59)*(9.984)^x
-------------
Another way to proceed:
Pick two points from your data set:
(100,14.1) and (35,37)
The form is y=ab^x
Each point gives a form:
14.1=ab^100
37=ab^35
------------
You need to solve for a and for b.
Divide the 1st by the 2nd to get:
(14.1/37)=b^65
Take the 65th root of both sides to get:
b=0.986....
Substitute that value into one of the equations to solve for "a", as follows:
37=a(0.986...)^35
37=a(0.608...)
EQUATION:
y=(0.608..)(0.986...)^x
---------------------------
You will get a little different equation if you pick
different points, but you will get an exponential equation.
--------
Hope this helps.
Cheers,
Stan H.