Table of Contents
ToggleThe laws of indices (sometimes called 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. The laws of indices can be used when numbers share the same base, or can be rewritten to share the same base.
The key index laws you need to know are:
These laws are not in the formula booklet for IB Mathematics HL, so you must memorize them.
You will need to be able to perform multiple steps using the laws of indices:
The index laws only work directly when terms have the same base, so be sure to change the base before applying any law if necessary. This might involve rewriting a number as a power of another number.
For example:
The laws of indices typically show up embedded in other questions — for instance, when working with logarithms or polynomials, or when simplifying complicated algebraic expressions. Watch out for opportunities to apply these laws to simplify or rearrange expressions in order to proceed with a problem more efficiently.
Here are 10 examples illustrating how to apply the laws of indices step by step.
Step 1: Expand the numerator using multiplication.
$ (3x^2)(2x^3 y^2) = 3 \times 2 \times x^{2+3} \times y^2 = 6x^5 y^2. $
Step 2: Divide by the denominator $6x^2 y$.
$ \frac{6x^5 y^2}{6x^2 y} = \frac{6}{6} \times x^{5-2} \times y^{2-1} = x^3 y. $
Final Answer: $\displaystyle x^3 y$
Step 1: Rewrite each factor as a power/fraction if needed.
For $(4x^2 y^{-4})^3$, apply the law $(abc)^n = a^n b^n c^n$: $ (4x^2 y^{-4})^3 = 4^3 x^{2 \times 3} y^{-4 \times 3} = 64 x^6 y^{-12}. $
For $(2x^3 y^{-1})^{-2}$, apply the power: $ (2x^3 y^{-1})^{-2} = 2^{-2} x^{-6} y^{2} = \frac{y^2}{4x^6}. $
Step 2: Combine them via multiplication.
$ 64 x^6 y^{-12} \times \frac{y^2}{4x^6} = 64 \times \frac{1}{4} \times x^6 \times x^{-6} \times y^{-12} \times y^2. $
Simplify step by step: $ 64 \div 4 = 16, \; x^{6 + (-6)} = x^0 = 1, \; y^{-12 + 2} = y^{-10}. $
So we get: $ 16 y^{-10}. $
Final Answer: $\displaystyle \frac{16}{y^{10}}$
Step 1: Apply the power of a power law: $(x^m)^n = x^{mn}$.
$ (x^3)^4 = x^{3 \times 4} = x^{12}. $
Step 2: Multiply $x^{12}$ by $x^{-5}$ using $x^m \times x^n = x^{m+n}$.
$ x^{12} \times x^{-5} = x^{12 + (-5)} = x^7. $
Final Answer: $\displaystyle x^7$
Step 1: Rewrite the fraction to separate numeric and variable factors.
$ = \frac{2}{4} \times \frac{x^2}{x^{-3}} \times \frac{y^3}{y^5}. $
Step 2: Simplify each part.
$ \frac{2}{4} = \frac{1}{2}, \quad \frac{x^2}{x^{-3}} = x^{2 - (-3)} = x^5, \quad \frac{y^3}{y^5} = y^{3 - 5} = y^{-2}. $
Combine them:
$ \frac{1}{2} \times x^5 \times y^{-2} = \frac{x^5}{2y^2}. $
Final Answer: $\displaystyle \frac{x^5}{2y^2}$
Step 1: Combine numeric coefficients and then combine like bases.
$
2 \times 4 = 8.
$
$
x^{3/2} \times x^{-1/2} = x^{3/2 + (-1/2)} = x^{2/2} = x^1 = x.
$
Step 2: So the product is:
$ 8x. $
Final Answer: $\displaystyle 8x$
Step 1: Expand the numerator using $(xy)^m = x^m y^m$:
$ (a^2 b)^3 = a^{2 \times 3} b^{1 \times 3} = a^6 b^3. $
Step 2: Divide by $a^{-1} b^4$:
$ \frac{a^6 b^3}{a^{-1} b^4} = a^{6 - (-1)} b^{3 - 4} = a^{7} b^{-1}. $ which is $ a^7 \frac{1}{b}. $
Final Answer: $\displaystyle \frac{a^7}{b}$
Step 1: Inside the parenthesis, note $y^{-2} = \frac{1}{y^2}$. So:
$ \frac{x^3}{y^{-2}} = x^3 \times y^2. $
Step 2: Now raise to the power $-1$:
$ (x^3 y^2)^{-1} = x^{-3} y^{-2}. $ Which is $ \frac{1}{x^3 y^2}. $
Final Answer: $\displaystyle \frac{1}{x^3 y^2}$
Step 1: Combine the multiplication first: $x^{3/4} \times x^{5/4} = x^{3/4 + 5/4} = x^{8/4} = x^2.$
Step 2: Now divide by $x^{2/4} = x^{1/2}.$
So $ x^2 \div x^{1/2} = x^{2 - 1/2} = x^{3/2}. $
Final Answer: $\displaystyle x^{3/2}$
Step 1: Multiply inside the radical first: $8x^6 \times 4x^3 = 32x^9.$
So the expression is $ \sqrt[3]{32x^9}. $
Step 2: Rewrite $32 = 2^5$, so $32x^9 = 2^5 x^9.$
Step 3: Taking the cube root: $\sqrt[3]{2^5 x^9} = 2^{5/3} x^{9/3} = 2^{5/3} x^3.$
Often you might leave $2^{5/3}$ as $2 \sqrt[3]{2^2}$ or $2^{1 + 2/3}$, but in index form $ 2^{5/3} $ is perfectly acceptable.
Final Answer: $\displaystyle 2^{5/3} x^3$