Chapter 3.3, Problem 85E

To determine

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

Answer to Problem 85E

  $\ln y=-\frac{1}{4}\ln x+\ln 2.5$

Explanation of Solution

Given:

The table:

x	1	2	3	4	5	6
y	2.5	2.102	1.9	1.768	1.672	1.597

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
$\ln y$	0.916	0.743	0.642	0.570	0.514	0.468

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 $\left(0,0.916\right)$ and $\left(0.693,0.743\right)$ , the slope of the line is:

  $\begin{array}{c}m=\frac{y{2}_{}-y_1}{x_2-x_1}\\ =\frac{0.743-0.916}{0.693-0}\\ \approx -\frac{1}{4}\end{array}$

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,

  $\ln y=-\frac{1}{4}\ln x+b$

Now, to find b , substitute any point $\left(x,y\right)$ from the original table, so lets substitute $\left(1,2.5\right)$ ,

  $\begin{array}{c}\ln 2.5=-\frac{1}{4}\ln 1+b\\ \ln 2.5=-\frac{1}{4}\left(0\right)+b\\ \ln 2.5=b\end{array}$

Thus, the equation of line is:

  $\ln y=-\frac{1}{4}\ln x+\ln 2.5$

Conclusion:

The logarithmic equation that relates y to x is $\ln y=-\frac{1}{4}\ln x+\ln 2.5$ .