Logarithmic Laws


$$ \log_{a}{(xy)} = \log_{a}{x} + \log_{a}{y} $$
$\textit{Proof}$
Let $A=\log_{a}{x}$ and $B=\log_{a}{y}$.
Then $a^A = x$ and $a^B=y$.
\( \begin{align}
a^A \times a^B&= xy \\
a^{A+B} &= xy \\
A+B &= \log_{a}{(xy)} \\
\therefore \log_{a}{x}+\log_{a}{y} &= \log_{a}{(xy)} \\
\end{align} \)

$$\log_{a}{\dfrac{x}{y}} = \log_{a}{x} – \log_{a}{y} $$
$\textit{Proof}$
Let $A=\log_{a}{x}$ and $B=\log_{a}{y}$.
Then $a^A = x$ and $a^B=y$.
\( \begin{align}
\dfrac{a^A}{a^B} &= \dfrac{a}{y} \\
a^{A-B} &= \dfrac{x}{y} \\
A-B &= \log_{a}{\dfrac{x}{y}} \\
\therefore \log_{a}{x} – \log_{a}{y} &= \log_{a}{\dfrac{x}{y}} \\
\end{align} \)

$$\log_{a}{x^n} = n\log_{a}{x}$$
$\textit{Proof}$
\( \begin{align}
\log_{a}{x^n} &= \log_{a}{(\overbrace{x \times x \times \cdots \times x}^{n})} \\
&= \overbrace{\log_{a}{x} + \log_{a}{x} + \cdots + \log_{a}{x}}^{n} \\
&= n\log_{a}{x} \\
\therefore \log_{a}{x^n} &= n\log_{a}{x} \\
\end{align} \)

$$\log_{a}{1} = 0$$
$\textit{Proof}$
\( \begin{align}
a^0 &= 1 \\
\therefore 0 &= \log_{a}{1} \\
\end{align} \)

$$\log_{a}{a} = 1$$
$\textit{Proof}$
\( \begin{align}
a^1 &= a \\
\therefore 1 &= \log_{a}{a} \\
\end{align} \)

Please ensure the following as often many students made the following mistakes in the past.

$$ \begin{align}
\log_{a}{x} + \log_{a}{y} &\ne \log_{a}{(x+y)} \\
\log_{a}{x} – \log_{a}{y} &\ne \log_{a}{(x-y)} \\
\log_{a}{x} \times \log_{a}{y} &\ne \log_{a}{(xy)} \\
\dfrac{\log_{a}{x}}{\log_{a}{y}} &\ne \log_{a}{\dfrac{x}{y}} \\
\end{align}$$

It is importantto remember that each rule works only works if the base $a$ is the same for each term.

Example 1

Use the logarithmic laws to write $\log_{2}{3} + \log_{2}{7}$ as a single logarithm.

\( \begin{align} \displaystyle
\log_{2}{3} + \log_{2}{7} &= \log_{2}{(3 \times 7)} \\
&= \log_{2}{21} \\
\end{align} \)

Example 2

Use the logarithmic laws to write $\log_{4}{24} – \log_{4}{6}$ as an integer.

\( \begin{align} \displaystyle
\log_{4}{24} – \log_{4}{6} &= \log_{4}{\dfrac{24}{6}} \\
&= \log_{4}{4} \\
&= 1 \\
\end{align} \)

Example 3

Use the logarithmic laws to write $2\log_{5}{4} – 3\log_{5}{2}$ as a single logarithm.

\( \begin{align} \displaystyle
2\log_{5}{4} – 3\log_{5}{2} &= \log_{5}{4^2} – \log_{5}{2^3} \\
&= \log_{5}{16} – \log_{5}{8} \\
&= \log_{5}{\dfrac{16}{8}} \\
&= \log_{5}{2} \\
\end{align} \)

Example 4

Use the logarithmic laws to write $\log_{a}{24}$ in terms of $x$ and $y$ given $x=\log_{a}{2}$ and $y=\log_{a}{3}$.

\( \begin{align} \displaystyle
\log_{a}{24} &= \log_{a}{(2 \times 2 \times 2 \times 3)} \\
&= \log_{a}{(2^3 \times 3)} \\
&= \log_{a}{2^3} + \log_{a}{3} \\
&= 3\log_{a}{2} + \log_{a}{3} \\
&= 3x + y \\
\end{align} \)

Example 5

Simplify $\dfrac{\log_{5}{729}}{\log_{5}{27}}$.

\( \begin{align} \displaystyle
\dfrac{\log_{5}{729}}{\log_{5}{27}} &= \dfrac{\log_{5}{3^6}}{\log_{5}{3^3}} \\
&= \dfrac{6\log_{5}{3}}{3\log_{5}{3}} \\
&= \dfrac{6}{3} \\
&= 2 \\
\end{align} \)


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