Chapter 6.3, Problem 26E

To determine

The reason that why the logarithmic expressions ${\log }_2\left(-1\right)$ and ${\log }_{1}1$ cannot defined.

Answer to Problem 26E

The expression ${\log }_2\left(-1\right)$ and ${\log }_{1}1$ are not defined.

Explanation of Solution

Given information:

The given logerithmic expressions are: ${\log }_2\left(-1\right)$ and ${\log }_{1}1$ .

Formula Used:

Exponential form:

If the logarithm form is $c={\log }_{a}b$ then its exponential form is $c=a^b$ .

Logarithmic division:

  ${\log }_{a}b=\frac{\log b}{\log a}$

Calculation:

The given logarithm form is ${\log }_2\left(-1\right)$ .

Let ${\log }_2\left(-1\right)=x$

Now, solve the expression for $x$ .

  $\begin{array}{c}{\log }_2\left(-1\right)=x\\ 2^x=-1\end{array}$

Since $2^x$ is a positive so it cannot be equal to $-1$ .

Similarly for the logarithm form ${\log }_{1}1$ .

Solve the expression, use the formula ${\log }_{a}b=\frac{\log b}{\log a}$ .

  ${\log }_{1}1=\frac{\log 1}{\log 1}$

It is known that $\log 1=0$ . Substitute this value in the above expression.

  $\begin{array}{c}{\log }_{1}1=\frac{\log 1}{\log 1}\\ =\frac{0}{0}\end{array}$

Thus, ${\log }_{1}1$ is undefined.

Hence, the expression ${\log }_2\left(-1\right)$ and ${\log }_{1}1$ are not defined.