Chapter 3, Problem 20PS

a.

To determine

Find the slope and $y$ −intercept of the line.

Answer to Problem 20PS

slope $m=\ln b$

Intercept $c=\ln a$

Explanation of Solution

Calculation:

Consider the equation,

  $y=a{b}^x$

Take the natural log of each side,

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

Now, if the equation of a straight line is expressed as $Y=mX+c$ , as the value $X$ is plotted along $x-axis$ and $Y$ is plotted along $y-axis$ , then the slope is $m$ and the intercept is $c$ .

So, consider,

  $\begin{array}{l}Y=\ln y\\ X=x\end{array}$

So, at the equation $\ln y=x\ln b+\ln a$ , slope is:

  $m=\ln b$

Intercept is,

  $c=\ln a$

Hence, slope is $m=\ln b$

Intercept is, $c=\ln a$

b.

To determine

Find the slope and $y$ −intercept of the line.

Answer to Problem 20PS

slope $m=b$

Intercept $c=\ln a$

Explanation of Solution

Calculation:

Consider the equation,

  $y=a{x}^b$

Take the natural log of each side,

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

Now, if the equation of a straight line is expressed as $Y=mX+c$ , as the value $X$ is plotted along $x-axis$ and $Y$ is plotted along $y-axis$ , then the slope is $m$ and the intercept is $c$ .

So, consider,

  $\begin{array}{l}Y=\ln y\\ X=\ln x\end{array}$

So, at the equation $\ln y=x\ln b+\ln a$ , slope is:

  $m=b$

Intercept is,

  $c=\ln a$

Hence, slope is $m=b$

Intercept is, $c=\ln a$