Chapter 3.3, Problem 86E

To determine

To find: the logarithmic equation that relates y to x .

Answer to Problem 86E

  $\ln y=\frac{5}{2}\ln x+\ln 0.5$

Explanation of Solution

Given:

The table:

x	1	2	3	4	5	6
y	0.5	2.828	7.794	16	27.951	44.091

Concept Used:

Slope of a line:

The slope of line passing through two point $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$ , $x_1\ne x_2$

Slope intercept form of equation of line is $y=mx+b$ , where m is the slope of the line and $\left(0,b\right)$ is the y -intercept of the line.

Construct a new table by taking the natural logarithms of the x and y values in the given table.

$\ln x$	0	0.693	1.099	1.386	1.609	1.792
$\ln y$	-0.693	1.040	2.053	2.773	3.330	3.786

Now plot each point $\left(\ln x,\ln y\right)$ to make sure they lie on a line.

  

Now, choose any two points to determine the slope of the line.

So, using the points $\left(0,-0.693\right)$ and $\left(0.693,1.040\right)$ , the slope of the line is:

  $\begin{array}{c}m=\frac{y{2}_{}-y_1}{x_2-x_1}\\ =\frac{1.040-\left(-0.693\right)}{0.693-0}\\ =\frac{1.733}{0.693}\\ \approx \frac{5}{2}\end{array}$

Now, from the slope-intercept form of line, the equation of line is:

  $Y=mX+b$

Where $Y=\ln y$ and $x=\ln x$

That is, equation of line is,

  $\ln y=\frac{5}{2}\ln x+b$

Now, to find b , substitute any point $\left(x,y\right)$ from the original table, so lets substitute $\left(1,0.5\right)$ ,

  $\begin{array}{c}\ln 0.5=\frac{5}{2}\ln 1+b\\ \ln 0.5=\frac{5}{2}\left(0\right)+b\\ \ln 0.5=b\end{array}$

Thus, the equation of line is:

  $\ln y=\frac{5}{2}\ln x+\ln 0.5$

Conclusion:

The logarithmic equation that relates y to x is $\ln y=\frac{5}{2}\ln x+\ln 0.5$ .