Logarithm (Laws Of Logarithm) - SS1 Mathematics Lesson Note

Logarithm is the exponent or power to which a base must be raised to yield a given number. Mathematically, \(\mathbf{\log}_{\mathbf{a}}\mathbf{b}\mathbf{= c}\) is true if \(\mathbf{a}^{\mathbf{c}}\mathbf{= b}\). In \(\mathbf{\log}_{\mathbf{a}}\mathbf{b}\mathbf{= c}\), \(\mathbf{a}\) is the base and \(\mathbf{c}\) is the exponent.

There are two commonly used logarithms based on their bases: common logarithms (\(\log_{10}N\ or\ \log N\)) make use of a base of \(10\) whilst natural logarithms (\(\log_{e}N\ or\ \ln N\)) make use of the Napierian base, \(e \approx 2.71\)

Given, \(\log_{10}100 = 2\); because \(10^{2} = 100\).

Product: \(\log_{a}{(MN) = \ \log_{a}M + \ \log_{a}N}\)

Example \(\log_{10}{(100 \times 1000)} = \ \log_{10}100 + \ \log_{10}1000\mathbf{=}2 + 3 = 5\)

Quotient: \(\log_{a}{(M \div N) = \ \log_{a}M - \ \log_{a}N}\)

Example \(\log_{3}{(81 \div 9)} = \ \log_{3}81 - \ \log_{3}9\mathbf{=}4 - 2 = 2\)

Raising to a power: \(\log_{a}{M^{P} = P\log_{a}M}\)

Example \(\log_{10}{(100)}^{2} = 2\log_{10}100 = 2(2) = 4\)

Roots: \(\log_{a}{\sqrt[n]{M} = \frac{\log_{a}M}{n}}\)

Example: \(\log_{10}\sqrt[3]{1000} = \ \frac{\log_{10}1000}{3} = \ \frac{3}{3} = 1\)

Zero: \(\log_{a}1 = 0\)

Identity: \(\log_{a}a = 1\)

Reciprocal: \(\log_{a}N = \ \frac{1}{\log_{N}a}\)

Change of Base: \(\log_{a}N = \ \frac{\log_{b}N}{\log_{b}a}\)

Example Given \(\mathbf{\log}_{\mathbf{10}}\mathbf{2}\mathbf{= 0.3010},\ \mathbf{\log}_{\mathbf{10}}\mathbf{3}\mathbf{= 0.4771}\ \)and\(\ \mathbf{\log}_{\mathbf{10}}\mathbf{7}\mathbf{= 0.8451}\), evaluate:

(a) \(\log_{10}6\)

(b) \(\log_{10}{(\frac{14}{3}})\)

Solution (a) \(\log_{10}6 = \ \log_{10}(2 \times 3) = \ \log_{10}2 + \ \log_{10}3 = \ 0.3010 + \ 0.4771 = 0.7781\)

(b) \(\log_{10}{(\frac{14}{3}}) = \log_{10}14 - \ \log_{10}3 = \ \log_{10}(2 \times 7) - \log_{10}3 = \ \log_{10}2 + \ \log_{10}7 - \ \log_{10}3 = (0.3010 + 0.8451) - \ 0.4771 = \ \ 0.6690\)