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.

Example 2

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

Example 3

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

Example 4

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

Example 5

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