Chapter 4.4, Problem 4E

a.

To determine

To define: The expended form of log $\left(\frac{x^{2}y}{z}\right)$ is to get.

Answer to Problem 4E

The logarithmic form of equation $2\log x+\log y-\log z$

Explanation of Solution

Given information:

The expended form of log $\left(\frac{x^{2}y}{z}\right)$ is .

The standard-form equation of logarithm is with the help of the laws of log.

It is also known as standard-form of logarithm.

The equation is ; log $\left(\frac{x^{2}y}{z}\right)$

Apply the low 2 of logarithm. That is;

  $\begin{array}{c}{\log }_a\left(\frac{A}{B}\right)={\log }_{a}A-{\log }_{a}B\\ \log \left(\frac{x^{2}y}{z}\right)=\log \left(x^{2}y\right)-\log z\\ =\log x^2+\log y-\log z\end{array}$

Apply low 1 of logarithm. That is

$\begin{array}{c}{\log }_a\left(AB\right)={\log }_{a}A+{\log }_{a}B\\ =\log x^2+\log y-\log z\\ =2\log x+\log y-\log z\end{array}$

b.

To determine

To define: The logarithm to get of equation $2\log x+\log y-\log z=\log x^2+\log y-\log z$

Answer to Problem 4E

The logarithmic form of equation $\log \left(\frac{x^{2}y}{z}\right)$

Explanation of Solution

Given information:

The expended form of logaritm. $2\log x+\log y-\log z=\log x^2+\log y-\log z$

The standard-form equation of logarithm is with the help of the laws of log.

It is also known as standard-form of logarithm.

The equation is ; $2\log x+\log y-\log z=\log x^2+\log y-\log z$

Apply the low 3 of logarithm. That is;

  $\begin{array}{c}{\log }_a\left(A^c\right)=c{\log }_{a}A\\ 2\log x+\log y-\log z=\log x^2+\log y-\log z\\ =\log \left(x^{2}y\right)-\log z\end{array}$

Apply low 1 of logarithm. That is

$\begin{array}{c}{\log }_a\left(AB\right)={\log }_{a}A+{\log }_{a}B\\ =\log \left(x^{2}y\right)-\log z\\ =\log \left(\frac{x^{2}y}{z}\right)\end{array}$