Chapter 3.4, Problem 58E

Solution

To determine: A given statement as true or false.

The given statement false.

Given data: The logarithm of a positive number is positive.

Method/Formula used:

The logarithm of a number x at the base a is denoted by ${\log }_{a}x$ , that in index form can be written as,

  $\begin{array}{c}{\log }_{a}x=p\\ \Rightarrow a^p=x\end{array}$

Explanation:

Let $x$ be a positive number, and ${\log }_{a}x=p\left(a>0\right)$ then index form of ${\log }_{a}x=p$ is $a^p=x$ . Now, ${\log }_{a}x$ is negative, i.e., $p$ is a negative number, then

  $\begin{array}{l}{a}^{-p}=x\\ {\left(a^{-1}\right)}^p=x\\ \Rightarrow {\left(\frac{1}{a}\right)}^p=x\leftarrow a^{-1}=\frac{1}{a}\end{array}$

Then by the definition of log

  $\begin{array}{l}{\left(\frac{1}{a}\right)}^p=x\\ {\log }_{1/a}x=p\end{array}$

That shows ${\log }_{1/a}x$ is always a positive number.

Moreover, let for two positive numbers a and x , ${\log }_{a}x<0$ , say ${\log }_{a}x=-k$ , then

  $\begin{array}{l}{\log }_{a}x=-k\\ a^{-k}=x\\ {\left(\frac{1}{a}\right)}^k=x\end{array}$

Taking log of both sides

  $\begin{array}{l}{\log }_a{\left(\frac{1}{a}\right)}^k=\log x\\ k{\log }_a\left(\frac{1}{a}\right)={\log }_{a}x\\ k\left({\log }_{a}1-{\log }_{a}a\right)={\log }_{a}x\leftarrow {\log }_{a}1=0\\ k=-\frac{{\log }_{a}x}{{\log }_{a}a}\end{array}$

Thus, k is a negative quantity, say, $k=-m\left(m>0\right)$ , then

  $\begin{array}{c}{\log }_{a}x=-\left(-m\right)\\ =m\\ \Rightarrow {\log }_{a}x>0\end{array}$

Therefore, the logarithm of a positive number is positive.

But consider the case: $\frac{1}{x}$ is a positive number, take log of it,

  $\begin{array}{c}\log \left(\frac{1}{x}\right)=\log 1-\log x\\ =-\log x\\ \Rightarrow \log \left(\frac{1}{x}\right)<0\end{array}$

Therefore, the logarithm of a positive number is not always positive.

Note: log of a negative number is undefined.