Hi guys i am trying to use the normal equations to fit some data to find values a and b for the following equation.
$$y=a*b^x$$
The data was given to me below and this is what i am currently using
My attempt at the question can be seen below
$$log y = log(a*b^x)$$ $$log y = log a +logb^x$$ $$log y = log a + x log b$$
$$Y = log y, A = log a, B=log b$$
$$Y = A + Bx$$
Based on the data given i got the following normal equations since it is linear form
$$7a_0 +3.9a_1 = 66.7$$ $$3.9a_0 - 2.95a_1 = 48.85 $$
Solving the simultaneous equations gives the following values for $$a_0 = 1.12, a_1=15.09 $$
Therefore we re substitute to get back the actual values $$log a = 1.12$$ $$a = e^{1.12}$$ $$log b = B = 15.09$$ $$b = e^{15.09} = 3576874.742$$
Therefore the equation is of the form
$$y = 3.605*3576874.742^x$$
Therefore i know something is wrong with the equation but i am not sure what it is can anyone help to get the values for a and b and show me what i am doing incorrectly?