Chapter 3.3, Problem 88E

a)

To determine

To graph:

The given two functions on the same viewing rectangle.

Answer to Problem 88E

The required graph is:

  

Explanation of Solution

Given:

Two logarithmic functions:

  $y_1=\frac{1}{2}\ln x-\ln \left(x+2\right),y_2=\ln \left(\frac{\sqrt{x}}{x+2}\right)$

Calculation:

Upon entering the given two equations on the TI-84 Plus calculator and then graphing them, we get the required graph as shown below:

  

b)

To determine

To create:

A table of values for the given functions and the graphs using the ‘table’ feature of the calculator.

Answer to Problem 88E

The required table is:

  

Explanation of Solution

Given:

Two logarithmic functions:

  $y_1=\frac{1}{2}\ln x-\ln \left(x+2\right),y_2=\ln \left(\frac{\sqrt{x}}{x+2}\right)$

Upon using the ‘table’ feature of the calculator, we get the tables for the given two functions as shown below:

  

c)

To determine

To conclude:

What do the graphs in part (a) and tables in part (b) conclude about the given two functions? Verify the conclusion algebraically.

Answer to Problem 88E

Both the functions have the same graph and the same table. Hence both the functions are equivalent to each other.

Explanation of Solution

Given:

Two logarithmic functions:

  $y_1=\frac{1}{2}\ln x-\ln \left(x+2\right),y_2=\ln \left(\frac{\sqrt{x}}{x+2}\right)$

From the graphs in part (a) and the tables in part (b), we can see that both the functions are equivalent to each others.

Algebraic verification:

Upon simplifying the first function using properties of logarithm, we get the second function.

  $\begin{array}{l}{y}_1=\frac{1}{2}\ln x-\ln \left(x+2\right)\\ y_1=\ln x^{\frac{1}{2}}-\ln \left(x+2\right)\left(\text{Exponent property of logarithms}\right)\\ y_1=\ln \sqrt{x}-\ln \left(x+2\right)\left(\text{Upon rewriting the exponent into radical form}\right)\\ y_1=\ln \frac{\sqrt{x}}{x+2}\left(\text{Addition property of logarithm}\right)\\ y_1=y_2\end{array}$

Hence, we have verified algebraically that both the functions are the same.