Chapter 3.5, Problem 73E

Solution

To justify: Justify the conclusion made about power regression in the three types of logarithmic re-expression.

Given information: Given the three types of logarithmic re-expression.

Explanation:

Let assume a set of data points $(x,y)$ , where $x>0$ and $y>0$ for all data.

Let define a new set of data points $(u,v)$ , where the $u=\ln x$ and $v=\ln y$ , for each corresponded point.

Assume that the points $(u,v)$ are a good fit for a linear regression equation, that is, they are arranged in a straight line in a scatter plot, or close enough to a straight line for a linear regression to be a viable option. Then find a linear regression equation in the form,

  $v=au+b$

Where $a$ and $b$ are regression coefficients such that $a\ne 0$ .

As set $u=\ln x$ and $v=\ln y$ , make that substitution in the regression equation, obtained-

  $\ln y=a\ln x+b$

Then apply the exponential function to both sides of the equation,

Raise $e$ to the power of whatever is on either side of the equation:

  $\begin{array}{l}{e}^{\ln \left(y\right)}=e^{a\ln x+b}\\ y=e^{a\ln x+b}\end{array}$

Apply exponential properties to the right-hand side:

  $y\text{}=e^{a\ln x}\cdot e^b$ (Product of powers)

  $y={(e^{\ln x})}^a\cdot e^b$ (Power of a power)

  $y=x^a\cdot e^b$ (Definition of the natural logarithm)

Finally, let $c=e^b$and rewrite the equation as:

  $y=c{x}^a$

Which is a power regression model for the original data point $(x,y)$ .