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Section 9.5 Derivations of the Properties of Logarithms

In this section we will derive several properties of logarithms. In the next section we will see several examples and encounter several exercises where the properties are applied to logarithmic expressions.

For each of the stated properties, In each expression, \(b\text{,}\) \(x\text{,}\) \(A\text{,}\) and \(B\) all represent positive real numbers and \(b \neq 1\text{.}\) The variable \(n\) can represent any real number.

\(\highlight{\log_{b}(AB)=\log_{b}(A)+\log_{b}(B)}\).

This property is the companion of the product rule of exponents:

\begin{equation*} b^x \cdot b^y=b^{x+y}. \end{equation*}

Specifically, suppose that

\begin{equation*} \log_b(A)=x\,\,\text{and}\,\,\log_b(B)=y. \end{equation*}

It immediately follows that

\begin{equation*} b^x=A\,\,\text{and}\,\,b^y=B. \end{equation*}

We now have the following.

\begin{align*} AB\amp=b^xb^y\\ \amp=b^{x+y} \end{align*}

Since \(AB=b^{x+y}\text{,}\) it follow that \(\log_b(AB)=x+y\text{.}\) Putting it all together we reach our conclusion.

\begin{align*} \log_b(AB)\amp=x+y\\ \amp=\log_b(A)+\log_b(B) \end{align*}
\(\highlight{\log_{b}\left(\frac{A}{B}\right)=\log_{b}(A)-\log_{b}(B)}\).

This property is the companion of the quotient rule of exponents:

\begin{equation*} \frac{b^x}{b^y}=b^{x-y}. \end{equation*}

Specifically, suppose that

\begin{equation*} \log_b(A)=x\,\,\text{and}\,\,\log_b(B)=y. \end{equation*}

It immediately follows that

\begin{equation*} b^x=A\,\,\text{and}\,\,b^y=B. \end{equation*}

We now have the following.

\begin{align*} \frac{A}{B}\amp=\frac{b^x}{b^y}\\ \amp=b^{x-y} \end{align*}

Since \(\frac{A}{B}=b^{x-y}\text{,}\) it follow that \(\log_b\left(\frac{A}{B}\right)=x-y\text{.}\) Putting it all together we reach our conclusion.

\begin{align*} \log_b\left(\frac{A}{B}\right)\amp=x-y\\ \amp=\log_b(A)-\log_b(B) \end{align*}
\(\highlight{\log_{b}\left(A^{n}\right)=n \cdot \log_{b}(A)}\).

This property is the companion of the power-to-a-power rule of exponents:

\begin{equation*} \left(b^x\right)^n=b^{xn} \end{equation*}

Specifically, suppose that

\begin{equation*} \log_b(A)=x. \end{equation*}

It immediately follows that

\begin{equation*} b^x=A. \end{equation*}

We now have the following.

\begin{align*} A^n\amp=\left(b^x\right)^n\\ \amp=b^{nx} \end{align*}

Since \(A^n=b^{nx}\text{,}\) it follow that \(\log_b\left(A^n\right)=nx\text{.}\) Putting it all together we reach our conclusion.

\begin{align*} \log_b\left(A^n\right)\amp=nx\\ \amp=n \cdot \log_b(A) \end{align*}
\(\highlight{\log_{b}(b)=1\,\,\text{and}\,\,\log_{b}(1)=0\,\,\text{and}\log_{b}(b^x)=x}\).

The fact that \(\log_b(b)=1\) is an immediate consequence of the fact that \(b^1=b\text{.}\)

The fact that \(\log_b(1)=0\) is an immediate consequence of the fact that \(b^0=1\text{.}\)

The fact that

\begin{equation*} \log_\highlightr{b}(\highlight{b^x})=\highlightg{x} \end{equation*}

is an immediate consequence of the fact that

\begin{equation*} \highlightr{b}^\highlightg{x}=\highlight{b^x}. \end{equation*}
\(\highlight{b^{\log_{b}(A)}=A}\).

The fact that

\begin{equation*} \highlightr{b}^{\highlight{\log_{b}(A)}}=\highlightg{A} \end{equation*}

is an immediate consequence of the fact that

\begin{equation*} \log_\highlightr{b}(\highlightg{A})=\highlight{\log_{b}(A)}. \end{equation*}
\(\highlight{\log_b(x)=\frac{\log(x)}{\log(b)}\,\,\text{(This property is called "Change of Base.")}}\).

We begin by assuming that

\begin{equation*} \log_b(x)=n \end{equation*}

from which it follows that

\begin{equation*} b^n=x\text{.} \end{equation*}

We now equate the log-base-ten of the last two expressions and apply the rule \(\log\left(b^{n}\right)=n \cdot \log(b)\text{.}\) We solve the resultant equation for \(n\text{.}\)

\begin{align*} b^n\amp=x\\ \log\left(b^n\right)\amp=\log(x)\\ n \cdot \log(b) \amp=\log(x)\\ \divideunder{n \cdot \log(b)}{\log(b)}\amp=\divideunder{\log(x)}{\log(b)}\\ n\amp=\frac{\log(x)}{\log(b)} \end{align*}

Our conclusion is at hand, because our starting point was the assumption that \(\log_b(x)=n\text{.}\) Putting the pieces together we have the following.

\begin{align*} \log_b(x)\amp=n\\ \amp=\frac{\log(x)}{\log(b)} \end{align*}

Note

I used a base of ten on the right side of the stated change of base formula. I could have used any base (the same in the numerator and denominator). I chose to use base 10 to minimize symbols and because most scientific calculators afford easy access to a log-base-10 key.

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