Express the following data in the form: $y-2=k(1+x)^n$

Question

Given the following five data: (4,2.45), (8,2.60), (15,2.80), (19,2.89) and (24,3.00); show that there is a relation of the form: $y-2=k(1+x)^n$; and find approximate values for $k$ and $n$.

First I did this out on graph paper, chose two extreme points and calculated the function to be approx: $$y=0.028x+2.343$$

Then I inputted the data into Geogebra and it gave me a best fit function of: $$y=0.03x+2.37$$

I am now satisfied with my attempt at approximating the function.

Then I took Geogebra's function and tried to change it to the form: $$y-2=k(1+x)^n$$

The nearest I got was: $$y-2=0.03x+0.37$$

I'm tempted to divide the right hand side by $0.03$ which would have given me a $k$ value of $0.03$ and an $n$ value of $1$.

But the book gives solutions of: $k=0.2$ and $n=0.5$

How would one have gotten this result?

If you sub the book's values in, then you do in fact get a function that approximately agrees with Geogebra's function, but this new function only exists for $x>-1$

Also, why in the world would one choose to express the function in this form with the $x$ domain limited like that?

Answer 1

Well you found an equation of the provided form with $n=1$.

However, to generally approach this problem where you don't have a linear relationship you can do the following.

• Subtract 2 from $y$
• Take logarithms of both sides

this will lead to $$ \ln(y-2) = \ln k + n\ln(1+ x) $$ this is a function you can plot as a linear relationship where $n$ is the gradient of the resultant log plot. Thus if the gradient is 1 then so is the power in the original form you are trying to match.