Before you begin
You should be comfortable multiplying and dividing numbers and simplifying fractions.
The intuition
Writing a 3 a^3 a 3 means a × a × a a \times a \times a a × a × a — three copies of a a a multiplied together. The laws of indices are shortcuts that follow from this definition.
Definitions
Base : the number being raised to a power.
Index (exponent) : the number of times the base is multiplied.
a 0 = 1 a^0 = 1 a 0 = 1 (for a ≠ 0 a \neq 0 a = 0 ).
a − n = 1 a n a^{-n} = \frac{1}{a^n} a − n = a n 1 .
a 1 / n = a n a^{1/n} = \sqrt[n]{a} a 1/ n = n a (the n n n -th root).
The laws of indices
Multiplication : a m × a n = a m + n a^m \times a^n = a^{m+n} a m × a n = a m + n
Division : a m ÷ a n = a m − n a^m \div a^n = a^{m-n} a m ÷ a n = a m − n
Power of a power : ( a m ) n = a m n (a^m)^n = a^{mn} ( a m ) n = a mn
Power of a product : ( a b ) n = a n b n (ab)^n = a^n b^n ( ab ) n = a n b n
Power of a quotient : ( a b ) n = a n b n \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} ( b a ) n = b n a n
Negative index : a − n = 1 a n a^{-n} = \frac{1}{a^n} a − n = a n 1
Fractional index : a m / n = ( a n ) m a^{m/n} = \left(\sqrt[n]{a}\right)^m a m / n = ( n a ) m
Step-by-step method
Identify the base. The laws only apply when the base is the same.
Decide which law matches the operation.
Apply the law to combine or simplify the indices.
Give the answer with a single index where possible.
Worked examples
Foundation example
Simplify x 5 × x 3 x^5 \times x^3 x 5 × x 3 .
Same base, multiplication: add the indices.
x 5 × x 3 = x 5 + 3 = x 8 x^5 \times x^3 = x^{5+3} = x^8 x 5 × x 3 = x 5 + 3 = x 8
Developing example
Simplify 6 a 7 b 2 2 a 3 b 5 \frac{6a^7 b^2}{2a^3 b^5} 2 a 3 b 5 6 a 7 b 2 and write your answer with positive indices.
Deal with numbers and each letter separately.
6 a 7 b 2 2 a 3 b 5 = 3 ⋅ a 7 − 3 ⋅ b 2 − 5 = 3 a 4 b − 3 = 3 a 4 b 3 \frac{6a^7 b^2}{2a^3 b^5} = 3 \cdot a^{7-3} \cdot b^{2-5} = 3a^4 b^{-3} = \frac{3a^4}{b^3} 2 a 3 b 5 6 a 7 b 2 = 3 ⋅ a 7 − 3 ⋅ b 2 − 5 = 3 a 4 b − 3 = b 3 3 a 4
Advanced example
Simplify ( 2 x − 2 ) 3 × ( x 5 ) 2 \left(2x^{-2}\right)^3 \times \left(x^5\right)^2 ( 2 x − 2 ) 3 × ( x 5 ) 2 with positive indices.
( 2 x − 2 ) 3 = 2 3 ⋅ x − 6 = 8 x − 6 \left(2x^{-2}\right)^3 = 2^3 \cdot x^{-6} = 8x^{-6} ( 2 x − 2 ) 3 = 2 3 ⋅ x − 6 = 8 x − 6
( x 5 ) 2 = x 10 \left(x^5\right)^2 = x^{10} ( x 5 ) 2 = x 10
8 x − 6 × x 10 = 8 x − 6 + 10 = 8 x 4 8x^{-6} \times x^{10} = 8x^{-6+10} = 8x^4 8 x − 6 × x 10 = 8 x − 6 + 10 = 8 x 4
Common mistakes
Different bases : x 3 × y 4 x^3 \times y^4 x 3 × y 4 cannot be simplified by adding indices.
Forgetting brackets : ( 2 x ) 3 = 8 x 3 (2x)^3 = 8x^3 ( 2 x ) 3 = 8 x 3 , not 2 x 3 2x^3 2 x 3 .
Mixing up signs : a − n a^{-n} a − n is a reciprocal, not a negative number.
Concept checklist
I can apply each of the seven laws.
I can convert between negative and fractional indices.
I can simplify expressions with mixed indices.
Practice questions
Simplify a 4 × a 9 a^4 \times a^9 a 4 × a 9 .
Simplify x 10 x 4 \frac{x^{10}}{x^4} x 4 x 10 .
Simplify ( y 3 ) 5 (y^3)^5 ( y 3 ) 5 .
Evaluate 5 − 2 5^{-2} 5 − 2 .
Simplify 8 x 6 y 3 4 x 2 y 7 \frac{8x^6 y^3}{4x^2 y^7} 4 x 2 y 7 8 x 6 y 3 with positive indices.
Simplify ( 3 a − 2 ) 2 × a 5 (3a^{-2})^2 \times a^5 ( 3 a − 2 ) 2 × a 5 with positive indices.
Challenge questions
Simplify ( 2 x 3 ) 4 x 5 ⋅ x 2 \frac{(2x^3)^4}{x^5 \cdot x^2} x 5 ⋅ x 2 ( 2 x 3 ) 4 with positive indices.
Show that 16 − 3 / 4 = 1 8 16^{-3/4} = \frac{1}{8} 1 6 − 3/4 = 8 1 .
Frequently asked questions
What is a negative index?
A negative index means the reciprocal of the positive power: a − n = 1 a n a^{-n} = \frac{1}{a^n} a − n = a n 1 .
Why is a 0 = 1 a^0 = 1 a 0 = 1 ?
Using the division law, a m ÷ a m = a m − m = a 0 a^m \div a^m = a^{m-m} = a^0 a m ÷ a m = a m − m = a 0 . But a m a m = 1 \frac{a^m}{a^m} = 1 a m a m = 1 , so a 0 = 1 a^0 = 1 a 0 = 1 (for a ≠ 0 a \neq 0 a = 0 ).
Standard form and very large or small numbers
Exponential functions and logarithms
Understanding surds and simplifying radicals
Strengthen your problem-solving
Concept enrichment and approved competition preparation with Dr. Pankaj Jha.
Book a Consultation SelexIQ provides supplementary mathematics enrichment, advanced problem-solving development and preparation for approved mathematics competition programmes. It does not provide or replace school-curriculum instruction.