Chapter 4.4, Problem 73E

To determine

To describe:The error in the expression.

  $\begin{array}{c}\log 0.1\prec 2\log 0.1\\ =\log {\left(0.1\right)}^2\\ =\log 0.01\\ \log 0.1\prec \log 0.01\\ 0.1\prec 0.01\end{array}$

Answer to Problem 73E

The correct expression is:

  $\begin{array}{c}\log 0.1\prec 2\log 0.1\\ 2\log 0.1=\log {\left(0.1\right)}^2\\ =\log 0.1\end{array}$

Explanation of Solution

Given information:

The expression:

  $\begin{array}{c}\log 0.1\prec 2\log 0.1\\ =\log {\left(0.1\right)}^2\\ =\log 0.01\\ \log 0.1\prec \log 0.01\\ 0.1\prec 0.01\end{array}$

Formula used:

Using the laws of logarithm:

  $2\log x=\log x^2$

Consider the expression.

  $\begin{array}{c}\log 0.1\prec 2\log 0.1\\ =\log {\left(0.1\right)}^2\\ =\log 0.01\\ \log 0.1\prec \log 0.01\\ 0.1\prec 0.01\end{array}$

This expression should be completely evaluate using the laws of logarithms.

In this expression $\log 0.1\prec 2\log 0.1$ is the right thing done here.

Then next $2\log 0.1\prec \log {\left(0.1\right)}^2$ , this is also a right step.

Then next:

  $\begin{array}{c}2\log 0.1\prec \log {\left(0.1\right)}^2\\ =\log 0.1\end{array}$

Once this has been done it can be found that there is not an error in the conversions and computations.

Then the problem is that the equation should be read $\log 0.1\succ 2\log 0.1$ .

After this the given problem is correct.