# Logarithmic Laws

$$\large \log_{a}{(xy)} = \log_{a}{x} + \log_{a}{y}$$
$\textit{Proof}$
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}

$$\large \log_{a}{\dfrac{x}{y}} = \log_{a}{x}-\log_{a}{y}$$
$\textit{Proof}$
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}

$$\large \log_{a}{x^n} = n\log_{a}{x}$$
$\textit{Proof}$
\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}

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

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

Please ensure the following, as many students have often made the following mistakes.

\large \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 important to remember that each rule works only 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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