1.2.2 Logarithms

Introduction to Logarithms

What are logarithms?

A logarithm is the inverse of an exponent
- If $a^{x} = b$ then $\log_{a} (b) = x$ where a > 0, b > 0, a ≠ 1
  - This is in the formula booklet
  - Your GDC will be able to use this function to solve equations involving exponents

Try to get used to ‘reading’ logarithm statements to yourself
- $\log_{a} (b) = x$ would be read as “the power that you raise $a$ to, to get $b$ , is $x$ ”
- So $\log_{5} 125 = 3$ would be read as “the power that you raise 5 to, to get 125, is 3”

Two important cases are:
- $\ln x = \log_{e} (x)$
  - Where e is the mathematical constant 2.718…
  - This is called the natural logarithm and will have its own button on your GDC
- $\log x = \log_{10} (x)$
  - Logarithms of base 10 are used often and so abbreviated to log x

Why use logarithms?

Logarithms allow us to solve equations where the exponent is the unknown value
- We can solve some of these by inspection
  - For example, for the equation 2^x = 8 we know that x must be 3
- Logarithms allow use to solve more complicated problems
  - For example, the equation 2^x = 10 does not have a clear answer
  - Instead, we can use our GDCs to find the value of $\log_{2} 10$

Exam Tip

Before going into the exam, make sure you are completely familiar with your GDC and know how to use its logarithm functions

Worked Example

Solve the following equations:

x = \log_{3} 27

ai-sl-1-1-2intro-to-logs-we-i

ii)

2^{x} = 21.4

, giving your answer to 3 s.f.

Laws of Logarithms

What are the laws of logarithms?

Laws of logarithms allow you to simplify and manipulate expressions involving logarithms
- The laws of logarithms are equivalent to the laws of indices
The laws you need to know are, given $a > 0$ :
- $\log_{a} x y = \log_{a} x + \log_{a} y$
  - This relates to $a^{x} \times a^{y} = a^{x + y}$
- $\log_{a} \frac{x}{y} = \log_{a} x {- \log}_{a} y$
  - This relates to $a^{x} \div a^{y} = a^{x - y}$
- $\log_{a} x^{m} = m \log_{a} x$
  - This relates to $(a^{x})^{y} = a^{x y}$
These laws are in the formula booklet so you do not need to remember them
- You must make sure you know how to use them

Laws of Logarithms Notes fig2

Useful results from the laws of logarithms

Given $a > 0, a \neq 1$
- $\log_{a} 1 = 0$
  - This is equivalent to $a^{0} = 1$
If we substitute b for a into the given identity in the formula booklet
- $a^{x} = b \Leftrightarrow \log_{a} b = x$ where $a > 0, b > 0, a \neq 1$
- $a^{x} = a \Leftrightarrow \log_{a} a = x$ gives $a^{1} = a \Leftrightarrow \log_{a} a = 1$
  - This is an important and useful result
Substituting this into the third law gives the result
- $\log_{a} a^{k} = k$
Taking the inverse of its operation gives the result
- $a^{\log_{a} x} = x$
From the third law we can also conclude that
- $\log_{a} \frac{1}{x} = - \log_{a} x$

Laws of Logarithms Notes fig3

These useful results are not in the formula booklet but can be deduced from the laws that are
Beware…
- … $\log_{a} (x + y) \neq \log_{a} x + \log_{a} y$
These results apply to $\ln x (\log_{e} x)$ too
- Two particularly useful results are
  - $\ln e^{x} = x$
  - $e^{\ln x} = x$
Laws of logarithms can be used to …
- simplify expressions
- solve logarithmic equations
- solve exponential equations

Exam Tip

Remember to check whether your solutions are valid
- log (x+k) is only defined if x > -k
- You will lose marks if you forget to reject invalid solutions

Worked Example

Write the expression

2 \log 4 - \log 2

in the form

\log k

, where

k \in ℤ

aa-sl-1-2-2-laws-of-logs-we-solution-part-a

b) Hence, or otherwise, solve

2 \log 4 - \log 2 = - \log \frac{1}{x}

aa-sl-1-2-2-laws-of-logs-we-solution-part-b

DP IB Maths: AI HL

Revision Notes

1.2.2 Logarithms

Introduction to Logarithms

What are logarithms?

Why use logarithms?

Exam Tip

Worked Example

Laws of Logarithms

What are the laws of logarithms?

Useful results from the laws of logarithms

Exam Tip

Worked Example

Author: Amber

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