Mathematics » Functions III » Exponential Functions

Laws of Logarithms

Laws of Logarithms

In earlier grades, we used the following exponential laws for working with exponents:

  • \({a}^{m} \times {a}^{n}={a}^{m+n}\)
  • \(\cfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}\)
  • \({(ab)}^{n}={a}^{n}{b}^{n}\)
  • \({\left(\cfrac{a}{b}\right)}^{n}=\cfrac{{a}^{n}}{{b}^{n}}\)
  • \({({a}^{m})}^{n}={a}^{mn}\)

where \(a > 0\), \(b > 0\) and \(m, n \in ℤ\).

The logarithmic laws are based on the exponential laws and make working with logarithms much easier.

Logarithmic laws:

  • \(\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0)\)
  • \(\log_{a}{x} = \cfrac{\log_{b}{x}}{\log_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)\)
  • \(\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\)
  • \(\log_{a}{\cfrac{x}{y}} = \log_{a}{x} – \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\)

The last two logarithmic laws in the list above are not covered in this section. They are discussed at the end of the tutorial and are included for enrichment only.

IMPORTANT: PROOFS ARE NOT REQUIRED FOR EXAMS

Logarithmic law:

\[\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0 )\]

\begin{align*} \text{Let } {\log}_{a}{x} &= m \ldots (1)\qquad (x > 0 ) \\ \therefore x &= a^{m} \\ \therefore (x)^{b} &= ( a^{m} )^{b} \\ \therefore x^{b} &= a^{bm} \\ \text{Change to logarithmic form: }\log_{a}(x^{b}) &= bm \\ \text{And subst}: \quad m &= {\log}_{a}{x} \\ \therefore {\log}_{a}{x^{b}} &= b{\log}_{a}{x} \end{align*}

In words: the logarithm of a number which is raised to a power is equal to the value of the power multiplied by the logarithm of the number.

Example

Question

Determine the value of \(\log_{3}{{27}^{4}}\).

Use the logarithmic law to simplify the expression

\begin{align*} \log_{3}{{27}^{4}} &= 4 \log_{3}{{27}}\\ &= 4\log_{3}{{3}^{3}} \\ &= (4 \times 3) \log_{3}{3}\\ &= 12 (1) \\ &= 12 \end{align*}

Write the final answer

\(\log_{3}{{27}^{4}} = 12\)

Special case:

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

The following is a special case of the logarithmic law \(\log_{a}{x^{b}} = b \log_{a}{x}\):

\begin{align*} \log_{a}{\sqrt[b]{x}} &= \log_{a}{{x}^{\frac{1}{b}}} \\ &= \cfrac{1}{b} \log_{a}{x} \\ &= \cfrac{\log_{a}{x}}{b} \end{align*}

Logarithmic law:

\({\log}_{a}x=\cfrac{{\log}_{b}{x}}{{\log}_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)\)

It is often necessary or convenient to convert a logarithm from one base to another base. This is referred to as a change of base.

\begin{align*} \text{Let } \quad \log_{a}{x} &= m \\ \therefore x &= a^{m} \\ \text{Consider the fraction: } \quad & \cfrac{\log_{b}{x}}{\log_{b}{a}} \\ \text{Substitute } x = a^{m}: \quad \cfrac{\log_{b}{x}}{\log_{b}{a}} &= \cfrac{\log_{b}{a^{m}}}{\log_{b}{a}} \\ &= m ( \cfrac{\log_{b}{a}}{\log_{b}{a}} ) \\ &= m (1) \\ \therefore \cfrac{\log_{b}{x}}{\log_{b}{a}}&= \log_{a}{x} \end{align*}

Special applications:

\begin{align*} (1) \qquad \log_{a}{x} &= \cfrac{\log_{x}{x}}{\log_{x}{a}} \\ \therefore \log_{a}{x} &= \cfrac{1}{\log_{x}{a}} \\ & \\ & \\ (2) \qquad \log_{a}{\cfrac{1}{x}} &= \log_{a}{x^{-1}} \\ \therefore \log_{a}{\cfrac{1}{x}} &= – \log_{a}{x} \end{align*}

 

Example

Question

Show: \(\log_{2}{8} = \cfrac{\log{8}}{\log{2}}\)

Simplify the right-hand side of the equation

\begin{align*} \text{RHS } &= \cfrac{\log{8}}{\log{2}} \\ &= \cfrac{\log{2^{3}}}{\log{2}} \\ &= 3 ( \cfrac{\log{2}}{\log{2}} ) \\ &= 3(1) \\ &= 3 \end{align*}

Simplify the left-hand side of the equation

\begin{align*} \text{LHS } &= \log_{2}{8} \\ &= \log_{2}{2^{3}} \\ &= 3 \log_{2}{2} \\ &= 3(1) \\ &= 3 \end{align*}

Write the final answer

We have shown that \(\log_{2}{8} = \dfrac{\log{8}}{\log{2}} = 3\).

Example

Question

If \(a = \log{2}\) and \(b = \log{3}\), express the following in terms of \(a\) and \(b\):

  1. \(\log_{3}{2}\)
  2. \(\log_{2}{\cfrac{10}{3}}\)

Use a change of base to simplify the expressions

  1. \begin{align*} \log_{3}{2} &= \cfrac{\log{2}}{\log{3}} \\ &= \cfrac{a}{b} \end{align*}
  2. \begin{align*} \log_{2}{\cfrac{10}{3}} &= \cfrac{\log{\cfrac{10}{3}}}{\log{2}} \\ &= \cfrac{\log{10} – \log{3}}{\log{2}} \\ &= \cfrac{1- b}{a} \end{align*}

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