Mathematics » Functions III » Enrichment: More On Logarithms

Laws of Logarithms II

Laws of Logarithms II

Logarithmic Law:

\[\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\]

\begin{align*} \text{Let } {\log}_{a}(x) = m & \quad \implies \quad x = a^{m} \ldots (1)\qquad (x > 0 )\\ \text{and }{\log}_{a}(y) = n & \quad \implies \quad x = a^{n} \ldots (2) \qquad (y > 0) \end{align*}\begin{align*} \text{Then } (1) \times (2): \quad x \times y &= a^{m} \times a^{n}\\ \therefore xy &= a^{m + n} \end{align*}

Now we change from the exponential form back to logarithmic form:

\begin{align*} \log_{a}{xy} &= m + n \\ \text{But } m = {\log}_{a}(x) & \text{ and } n = {\log}_{a}(y) \\ \therefore \log_{a}{xy} &= {\log}_{a}(x) +{\log}_{a}(y) \end{align*}

In words: the logarithm of a product is equal to the sum of the logarithms of the factors.

Example

Question

Simplify: \(\log{5} + \log{2} – \log{30}\)

Use the logarithmic law to simplify the expression

We combine the first two terms since the product of \(\text{5}\) and \(\text{2}\) is equal to \(\text{10}\), which is always useful when simplifying logarithms.

\begin{align*} \log{5} + \log{2} – \log{30} &= ( \log{5} + \log{2} ) – \log{30} \\ &= \log{( 5 \times 2)} – \log{30} \\ &= \log{10} – \log{30} \\ &= 1 – \log{30} \end{align*}

We expand the last term to simplify the expression further:

\begin{align*} &= 1 – \log{( 3 \times 10)} \\ &= 1 – ( \log{3} + \log{10} ) \\ &= 1 – ( \log{3} + 1 ) \\ &= 1 – \log{3} – 1 \\ &= – \log{3} \end{align*}

Write the final answer

\(\log{5} + \log{2} – \log{30} = – \log{3}\)

Logarithmic law:

\[\log_{a}{\cfrac{x}{y}} = \log_{a}{x} – \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\]

\begin{align*} \text{Let } {\log}_{a}(x) = m & \quad \implies \quad x = a^{m} \ldots (1) \qquad (x > 0)\\ \text{and }{\log}_{a}(y) = n & \quad \implies \quad y = a^{n} \ldots (2) \qquad (y > 0) \end{align*}\begin{align*} \text{Then } (1) \div (2): \quad \cfrac{x}{y} &= \cfrac{a^{m}}{a^{n}} \\ \therefore \cfrac{x}{y} &= a^{m – n} \end{align*}

Now we change from the exponential form back to logarithmic form:

\begin{align*} \log_{a}{\cfrac{x}{y}} &= m – n \\ \text{But } m = {\log}_{a}(x) & \text{ and } n = {\log}_{a}(y) \\ \therefore \log_{a}{\cfrac{x}{y}} &= {\log}_{a}(x) – {\log}_{a}(y) \end{align*}

In words: the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator.

Example

Question

Simplify: \(\log{40} – \log{4} + \log_{5}{\cfrac{8}{5}}\)

Use the logarithmic law to simplify the expression

We combine the first two terms since both terms have the same base and the quotient of \(\text{40}\) and \(\text{4}\) is equal to \(\text{10}\):

\begin{align*} \log{40} – \log{4} + \log_{5}{\cfrac{8}{5}} &= (\log{40} – \log{4} ) + \log_{5}{\cfrac{8}{5}} \\ &= (\log{\cfrac{40}{4}} ) + \log_{5}{\cfrac{8}{5}} \\ &= \log{10} + \log_{5}{\cfrac{8}{5}} \\ &= 1 + \log_{5}{\cfrac{8}{5}} \end{align*}

We expand the last term to simplify the expression further:

\begin{align*} &= 1 + ( \log_{5}{8} – \log_{5}{5} ) \\ &= 1 + \log_{5}{8} – 1 \\ &= \log_{5}{8} \end{align*}

Write the final answer

\(\log{40} – \log{4} + \log_{5}{\cfrac{8}{5}} = \log_{5}{8}\)

Useful summary:

  1. \(\log1=0\)

  2. \(\log10=1\)

  3. \(\log100=2\)

  4. \(\log1000=3\)

  5. \(\log{\cfrac{1}{10}}=-1\)

  6. \(\log{\text{0.1}}=-1\)

  7. \(\log{\text{0.01}}=-2\)

  8. \(\log{\text{0.001}}=-3\)

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