Logarithms | DP IB Maths: AI HL Revision Notes 2021

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Introduction to 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
  - The number a is called the base of the logarithm
  - 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”

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

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.

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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 $a, x, y > 0$ :
- $\log_{a} x y = \log_{a} x + \log_{a} y$
  - 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

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

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

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