Summary O Level Additional Mathematics Chapter on Logarithms and Exponential Functions

CHAPTER 5: LOGARITHMS AND EXPONENTIAL FUNCTIONS



Chapter objectives:

• Know and use the simple properties and graphs of logarithms and
exponential functions including ln𝑥 and 𝑒 T and graphs of 𝑘𝑒 •T + 𝑎
and 𝑘ln(𝑎𝑥 + 𝑏) where 𝑘, 𝑎, 𝑛 and 𝑏 are intergers.
• Know and use the laws of logarithms (including change of base of
logarithms)
• Solve equations of the form 𝑎T = 𝑏



A logarithm is a power. The logarithm of a number is the power to which
another number (the base) is raised to get that number. For example, the
logarithm to the base 10 of 100 (log/k100) is 2 because the power to
which the base (10) is raised to get 100 is two, i.e. 100 = 100 .

An exponent is the reverse of a logarithm e.g. log/k 100 = 2 → 100 = 100.

In general, 𝐥𝐨𝐠𝒂 𝒃 = 𝒙 → 𝒂𝒙 = 𝒃 and this relationship is important when
solving equations involving exponents and/or logarithms.



Special logarithms and exponents

log/k𝑥 is sometimes written lg 𝑥.

log à 𝑥, the natural logarithm is written ln 𝑥.
(𝑒 is the natural number and is equal to 2.7182818 … … .)



Laws of logarithms and exponential functions

87

,The laws of logarithms include but are not limited to:

1. log(𝐴𝐵 ) = log (𝐴) + log (𝐵)
Ä
2. log x y = log(𝐴) − log (𝐵)
Å
T)
3. log(𝐴 = 𝑥 log(𝐴)
/
4. log x y = −log (𝐴)
Ä
5. log 1 = 0
6. log Ä (𝐴) = 1

These laws of logarithms also apply to natural logarithms:

1. ln(𝐴𝐵 ) = ln 𝐴 + ln 𝐵
Ä
2. ln x y = ln 𝐴 − ln 𝐵
Å
T
3. ln(𝐴 ) = 𝑥 ln 𝐴
/
4. ln x y = − ln 𝐴
Ä
5. ln 1 = 0
6. ln 𝑒 = 1

These laws are important in solving the equations in the form 𝑎T = 𝑏



Example 5.1
/
Without using a calculator find the value of lg x y.
/k¥




SOLUTION
1
lg ‡ ˆ = − lg(10• )
10•
− lg(10• ) = −𝑛 lg(10)

lg = log/k
88

, → −𝑛 lg(10) = −𝑛 × 1
1
∴ lg ‡ ˆ = −𝑛
10•



Example 5.2

Solve the equation log “(2𝑥 ) + log “ (3𝑥 ) = 2



SOLUTION

log “(2𝑥 ) + log “ (3𝑥 ) = 2

→ log “ (2𝑥 × 3𝑥 ) = 2

→ log “ (6𝑥 0 ) = 2

2 = log “ 60

→ log “ (6𝑥 0 ) = log “ 60

The bases are equal

→ 6𝑥 0 = 60

→ 𝑥0 = 6

→ 𝑥 = ±√6

It is always important in questions involving logarithms to check back each
answer in the original equation to see if it makes sense in that context
bearing in mind that logarithms of negative numbers are non-existent. And
so the solution 𝑥 = −√6 is rejected since it gives the logarithm of a negative
number when substituted in the original equation.

∴ 𝑥 = √6
89