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SS1: MATHEMATICS - 1ST TERM

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  1. Number Base System I | Week 1
    6Topics
    |
    2 Quizzes
  2. Number Base System II | Week 2
    3Topics
    |
    1 Quiz
  3. Number Base System III | Week 3
    2Topics
    |
    1 Quiz
  4. Modular Arithmetic I | Week 4
    2Topics
  5. Modular Arithmetic II | Week 5
    2Topics
  6. Modular Arithmetic III | Week 6
    3Topics
    |
    1 Quiz
  7. Indices I | Week 7
    2Topics
  8. Indices II | Week 8
    1Topic
    |
    1 Quiz
  9. Logarithms I | Week 9
    3Topics
  10. Logarithms II
    1Topic
    |
    1 Quiz
Lesson Progress
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1) \( \scriptsize \log_a MN = \log_a M \: +\: \log_aN \\ \rightarrow \scriptsize a^x \: \times \: a^y \\ \scriptsize = a^{x + y} \)

2) \( \scriptsize \log_a \normalsize\left (\frac{M}{N} \right ) \scriptsize = \log_a M \: – \: \log_aN \\ \rightarrow \scriptsize a^x \div a^y \\ \scriptsize = a^{x – y}\)

3) \(\scriptsize \log_{a} \left ( M^{n} \right ) = n \log_{a} M \\ \scriptsize \rightarrow \left ( a^{x} \right )^{y} = a^{xy} \)

(for any base “a” > 0, where \( \scriptsize a \neq 1\))


  • Note that the three basic laws of logarithms are closely related to those of indices given earlier on.
  • Note that \( \frac {\log M}{\log N} \scriptsize \neq \log M \; – \log N \)⚠️ (Very Important)

Special Logarithms

4) \(\scriptsize \log_{a}a = 1 \\ \scriptsize\rightarrow a^1 = a \)

5) \(\scriptsize \log_{a}1 = 0 \\ \scriptsize \rightarrow a^0 \)

6) \(\scriptsize \log_{a}\normalsize\left (\frac{1}{a} \right ) \scriptsize = -1 \\ \scriptsize \rightarrow a^{-1} = \normalsize \frac{1}{a} \)

7) \(\scriptsize \log_{a}\normalsize\left (\frac{1}{x} \right ) \scriptsize = -\log_a x \)


Changing the Base of a Logarithm:

Note that it is possible to change the base of a given logarithm to a more convenient base.

Suppose we wanted to find the value of the expression \(\scriptsize \log_2(50)\) Since 50 is not a rational power of 2, it is difficult to evaluate this without a calculator.

However, most calculators only directly calculate logarithms in base-10 and base e. So in order to find the value of \(\scriptsize \log_2(50)\) we must change the base of the logarithm first.

Suppose  \(\scriptsize \log_{q} P = y, \; then \; q^y = P \)

Taking logs to base a of both sides of 

\(\scriptsize q^y = P \)

hence, \(\scriptsize \log_{a} q^y = \log_{a} P \)

i.e \(\scriptsize y \log_{a} q = \log_{a} P \)

i.e \(\scriptsize y = \normalsize \frac{\log_{a} P}{\log_{a} q} \)

\( \therefore \scriptsize \log_{q} P = \normalsize \frac{\log_{a} P}{\log_{a} q} \)

(where p, q are positive real numbers \( \neq \) 1).

If  a = P  then,

\( \scriptsize \log_{q} a = \normalsize \frac{\log_{a} a}{\log_{a} q} \)

\( \scriptsize \log_{q} a = \normalsize \frac{1}{\log_{a} q} \)

The change of base rule

The base of any logarithm can easily be changed by using the following rule:

\( \scriptsize \log_{b} (a) = \normalsize \frac{\log_x (a)}{\log_x (b)} \) See Example

Example: Evaluating \( {\color{Yellow}\scriptsize \log_2(50) }\)

If your goal is to find the value of a logarithm, change the base to 10 or e since these logarithms can be calculated on most calculators.

So let’s change the base of  \( \scriptsize \log_2 50 \; to \; 10 \)

Change of base rule \( \scriptsize \log_{b} (a) = \normalsize \frac{\log_x (a)}{\log_x (b)} \)

Using this rule, a = 50, b = 2, x = 10

\( \therefore \scriptsize \log_{2} (50) = \normalsize \frac{\log_{10} (50)}{\log_{10} (2)} \)

= \( \frac{\log (50)}{\log(2)} \)

Using a calculator this gives 5.644

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