Exponents & Logarithms (DP IB Maths: AI SL)

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Laws of indices (or index laws) allow you to simplify and manipulate expressions involving exponents
- An exponent is a power that a number (called the base) is raised to
- Laws of indices can be used when the numbers are written with the same base
The index laws you need to know are:
- ${(x y)}^{m} = x^{m} y^{m}$
- ${(\frac{x}{y})}^{m} = \frac{x^{m}}{y^{m}}$
- $x^{m} \times x^{n} = x^{m + n}$
- $x^{m} \div x^{n} = x^{m - n}$
- ${(x^{m})}^{n} = x^{m n}$
- $x^{1} = x$
- $x^{0} = 1$
- $\frac{1}{x^{m}} = x^{- m}$
These laws are not in the formula booklet so you must remember them

You will need to be able to carry out multiple calculations with the laws of indices
- Take your time and apply each law individually
- Work with numbers first and then with algebra
Index laws only work with terms that have the same base, make sure you change the base of the term before using any of the index laws
- Changing the base means rewriting the number as an exponent with the base you need
- For example, $9^{4} = (3^{2})^{4} = 3^{2 \times 4} = 3^{8}$
- Using the above can them help with problems like $9^{4} \div 3^{7} = 3^{8} \div 3^{7} = 3^{1} = 3$

Index laws are rarely a question on their own in the exam but are often needed to help you solve other problems, especially when working with logarithms or polynomials
Look out for times when the laws of indices can be applied to help you solve a problem algebraically

Simplify the following equations:

\frac{(3 x^{2}) (2 x^{3} y^{2})}{(6 x^{2} y)}

ai-sl-1-1-2-laws-of-indices-we-i

ii)

(4 x^{2} y^{- 4})^{3} (2 x^{3} y^{- 1})^{- 2}

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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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