Chapter 10.CR, Problem 1CR

To determine

(a)

To simplify:

The expression $\log 10+\ln e+{\log }_{2}2$.

Answer to Problem 1CR

Solution:

The simplified form of the given expression is 3.

Explanation of Solution

Approach:

For positive $b$ and $A$, $b\ne 1$.

$b^x=A$ means $x={\log }_{b}A$

Here $x$ is called the logarithm and $A$ is called the argument.

The result on logarithm is given as,

$\log a_{}a=1$

Calculation:

Consider the given expression,

$\log 10+\ln e+{\log }_{2}2$

Now, $\log 10$ is the exponent on 10, so it can be written as ${\log }_{10}10$ and $\ln e$ is the exponent on $e$, so it can be written as ${\ln }_{e}e$, $\log 2$ is the exponent on 2, so it can be written as ${\log }_{2}2$.

The given expression can be simplified as,

$\begin{array}{rll}\log 10+\ln e+{\log }_{2}2& =& {\log }_{10}10+{\ln }_{e}e+{\log }_{2}2\\ & =& 1+1+1\\ & =& 3\end{array}$

The simplified form of the given expression is 3.

To determine

(b)

To simplify:

The expression ${\log }_5\frac{25}{125}$.

Answer to Problem 1CR

Solution:

The simplified form of the given expression is $-1$.

Explanation of Solution

Approach:

For positive $b$ and $A$, $b\ne 1$.

$b^x=A$ means $x={\log }_{b}A$

Here $x$ is called the logarithm and $A$ is called the argument.

The results on logarithm are given as,

${\text{log}}_b\frac{x}{y}={\log }_{b}x-{\log }_{b}y\text{and}{\text{log}}_{a}a=1$

Also, ${\log }_{a}1=0$

Calculation:

Consider the given expression,

${\log }_5\frac{25}{125}$

The given expression can be simplified as,

$\begin{array}{rll}{\log }_5\frac{25}{125}& =& {\log }_5\frac{5^2}{5^3}\\ & =& {\log }_5\frac{1}{5}\end{array}$

By results on logarithm,

$\begin{array}{rll}{\log }_5\frac{1}{5}& =& {\log }_{5}1-{\log }_{5}5\\ & =& 0-1\\ & =& -1\end{array}$

The simplified form of the given expression is $-1$.