Learning Goal
Syllabus Requirement
Students need to learn the
laws of indices for all rational exponents, with the below point being highlighted in particular:

The equivalence of ??a^{\frac{m}{n}}?? and ??\sqrt[n]{a^{m}}??.
Study Notes
There are 7 basic laws of indices that should be learnt and understood. They are listed in the information box below:
The 7 basic laws of indices.
Law 1:
??x^{m} \times x^{n} = x^{(m+n)}??
Law 2:
??x^{m} \div x^{n} = x^{(mn)}??
Law 3:
??(x^{m})^{n} = x^{mn}??
Law 4:
??x^{0} = 1??
Law 5:
??x^{n} = \frac{1}{x^{n}}??
Law 6:
??x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}??
Law 7:
??x^{\frac{1}{n}} = \sqrt[n]{x}??
Note that Law 7 is actually just a special case of Law 6 (i.e. the case where ??m = 1??).
It is important to understand two terms, the base is the number, or expression that is being raised to a power or index. For example, in the term ??a^{b}??, the base is ??a?? and the power or index is ??b??.
Worked Examples
We'll now run through some examples of each of these rules in turn.
Law 1
Example 1
Question
Simplify the following expression ?? x^{2} \times x^{3} ??.
Show Solution
Worked Solution
??x^{2} \times x^{3} = x^{2 + 3} = x^{5}??
Example 2
Question
Simplify the following expression ?? x^{4} \times x^{3} \times x^{5} ??.
Show Solution
Worked Solution
??x^{4} \times x^{3} \times x^{5} = x^{4 + 3 + 5} = x^{12}??
Law 2
Example 3
Question
Simplify the following expression ?? x^{3} \div x^{2} ??.
Show Solution
Worked Solution
??x^{3} \div x^{2} = x^{3  2} = x^{1} = x??
Law 3
Example 4
Question
Simplify the following expression ?? (x^{2})^{5} ??.
Show Solution
Worked Solution
??(x^{2})^{5} = x^{2 \times 5} = x^{10} ??
Law 4
Example 5
Question
Simplify the following expression ?? x^{3} \times x^{3} ??.
Show Solution
Worked Solution
Apply Law 1 first, before applying Law 4:
?? x^{3} \times x^{3} = x^{3 + (3)} = x^{0} = 1 ??
Law 5
Example 6
Question
Write the following expression with only positive indices ?? x^{3} \times x^{3} ??.
Show Solution
Worked Solution
Apply Law 1 first, before applying Law 5:
?? x^{3} \times x^{3} = x^{(3) + (3)} = x^{6} = \frac{1}{x^6} ??
Law 6
Example 7
Question
Simplify the following expression ?? x^{\frac{4}{3}}??.
Show Solution
Worked Solution
??x^{\frac{4}{3}} = \sqrt[3]{x^{4}} = (\sqrt[3]{x})^{4}??
Either of the last two forms are acceptable.
Law 7
Example 8
Question
Simplify the following expression ?? x^{\frac{1}{3}}??.
Show Solution
Worked Solution
??x^{\frac{1}{3}} = \sqrt[3]{x}??
Exam Questions
The table below contains every exam question that has been asked on this topic, this includes normal papers from both January
and June sittings, International papers and Specimen papers.
To see an exam question and solution simply click on the load question icon (), the question will appear below the table and
the solution can be shown by clicking the "Show Solution" button that also appears.
The full exam paper and mark schemes are also available for download by clicking on the download icons in each
row of the table ().
Load Question 
Exam Board 
Subject 
Paper 
Year 
Month 
Module 
Question No. 
Parts 
Total Marks 
Exam Paper 
Mark Scheme 

Edexcel 
Maths 
Standard 
2005 
January 
Core 1 
1 
(a), (b) 
3 



Edexcel 
Maths 
Standard 
2005 
June 
Core 1 
1 
(a), (b) 
3 



Edexcel 
Maths 
Standard 
2007 
June 
Core 1 
2 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2008 
January 
Core 1 
2 
(a), (b) 
3 



Edexcel 
Maths 
Standard 
2009 
January 
Core 1 
1 
(a), (b) 
3 



Edexcel 
Maths 
Standard 
2009 
June 
Core 1 
2 
n/a 
3 



Edexcel 
Maths 
Standard 
2011 
January 
Core 1 
1 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2011 
June 
Core 1 
1 
(a), (b) 
3 



Edexcel 
Maths 
Standard 
2012 
June 
Core 1 
2 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2013 
January 
Core 1 
2 
n/a 
2 



Edexcel 
Maths 
Standard 
2013 
June 
Core 1 
3 
(a), (b) 
5 



Edexcel 
Maths 
International 
2013 
June 
Core 1 
5 
(a), (b) 
5 


Edexcel  Maths  Standard  2005  January  Core 1  Question 1
Question
1. (a)
Write down the value of ??16^{\frac{1}{2}}??. [1]
1. (b)
Find the value of ??16^{\frac{3}{2}}??. [2]
Hide Question
Show Solution
Worked Solution
1. (a)
Start by noting that the question asks you to write down the answer, which means it should not take very much, if any, working at all. Note that Law 7 states that:
$$x^{\frac{1}{n}} = \sqrt[n]{x}$$
So we can say that:
$$16^{\frac{1}{2}} = \sqrt{16} = 4$$
1. (b)
We know, from 1. (a) that ??16^{\frac{1}{2}} = 4?? and we also know from Law 6 that:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
And from Law 5 that:
$$x^{n} = \frac{1}{x^{n}}$$
Which means that:
$$16^{\frac{3}{2}} = \frac{1}{(\sqrt{16})^{3}} = \frac{1}{4^{3}} = \frac{1}{64}$$
Edexcel  Maths  Standard  2005  June  Core 1  Question 1
Question
1. (a)
Write down the value of ??8^{\frac{1}{3}}??. [1]
1. (b)
Find the value of ??8^{\frac{2}{3}}??. [2]
Hide Question
Show Solution
Worked Solution
1. (a)
Start by noting that the question asks you to write down the answer, which means it should not take very much, if any, working at all. Note that Law 7 states that:
$$x^{\frac{1}{n}} = \sqrt[n]{x}$$
So we can say that:
$$8^{\frac{1}{3}} = \sqrt[3]{8} = 2$$
1. (b)
We know, from 1. (a) that ??8^{\frac{1}{3}} = 2?? and we also know from Law 6 that:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
And from Law 5 that:
$$x^{n} = \frac{1}{x^{n}}$$
Which means that:
$$8^{\frac{2}{3}} = \frac{1}{(\sqrt[3]{8})^{2}} = \frac{1}{2^{2}} = \frac{1}{4}$$
Edexcel  Maths  Standard  2007  June  Core 1  Question 2
Question
2. (a)
Find the value of ??8^{\frac{4}{3}}??. [2]
2. (b)
Simplify ??\frac{15x^{\frac{4}{3}}}{3x}??. [2]
Hide Question
Show Solution
Worked Solution
2. (a)
Note that Law 6 states:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
So we can say that:
$$8^{\frac{4}{3}} = (\sqrt[3]{8})^{4} = 2^{4} = 16$$
2. (b)
Note that the denominator ??3x?? is made up of a factor of ??3?? and ??x?? both of which are also factors of the numerator. Lets start by dividing the
numerator and denominator by ??3??:
$$\frac{15x^{\frac{4}{3}}}{3x}$$
$$= \frac{15x^{\frac{4}{3}} \div 3}{3x \div 3}$$
$$ = \frac{5x^{\frac{4}{3}}}{x}$$
And then divide both numerator and denominator by ??x??:
$$\frac{5x^{\frac{4}{3}}}{x}$$
$$=\frac{5x^{\frac{4}{3}}\div x}{x \div x}$$
$$=\frac{5x^{\frac{4}{3}}\div x}{1}$$
$$=5x^{\frac{4}{3}}\div x$$
Now note that Law 2 states that:
$$x^{m} \div x^{n} = x^{(mn)}$$
So that we can write:
$$5x^{\frac{4}{3}}\div x$$
$$=5x^{\frac{4}{3}1}$$
$$=5x^{\frac{1}{3}}$$
Edexcel  Maths  Standard  2008  January  Core 1  Question 2
Question
2. (a)
Write down the value of ??16^{\frac{1}{4}}??. [1]
2. (b)
Simplify ??(16x^{12})^{\frac{3}{4}}??. [2]
Hide Question
Show Solution
Worked Solution
2. (a)
Note that Law 6 states:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
So we can say that:
$$16^{\frac{1}{4}} = \sqrt[4]{16} = 2$$
2. (b)
Note that Law 3 states:
$$(x^{m})^{n} = x^{mn}$$
And so we can write, noting that the power of ??\frac{3}{4}?? applies to both terms (i.e. the ??16?? and ??x^{12}?? within the brackets):
$$(16x^{12})^{\frac{3}{4}}$$
$$=16^{\frac{3}{4}} \times x^{12 \times \frac{3}{4}}$$
$$=16^{\frac{3}{4}} \times x^{\frac{36}{4}}$$
$$=16^{\frac{3}{4}} \times x^{9}$$
Now we can use Law 6:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
Which allows us to simplify the ??16^{\frac{3}{4}}?? term:
$$16^{\frac{3}{4}} \times x^{9}$$
$$= (\sqrt[4]{16})^{3} \times x^{9}$$
$$= 2^{3} \times x^{9}$$
$$= 8x^{9}$$
Edexcel  Maths  Standard  2009  January  Core 1  Question 1
Question
1. (a)
Write down the value of ??125^{\frac{1}{3}}??. [1]
1. (b)
Find the value of ??125^{\frac{2}{3}}??. [2]
Hide Question
Show Solution
Worked Solution
1. (a)
Start by noting that the question asks you to write down the answer, which means it should not take very much, if any, working at all. Note that Law 7 states that:
$$x^{\frac{1}{n}} = \sqrt[n]{x}$$
So we can say that:
$$125^{\frac{1}{3}} = \sqrt[3]{125} = 5$$
1. (b)
We know, from 1. (a) that ??125^{\frac{1}{3}} = 5?? and we also know from Law 6 that:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
And from Law 5 that:
$$x^{n} = \frac{1}{x^{n}}$$
Which means that:
$$125^{\frac{2}{3}} = \frac{1}{(\sqrt[3]{125})^{2}} = \frac{1}{5^{2}} = \frac{1}{25}$$
Edexcel  Maths  Standard  2009  June  Core 1  Question 2
Question
2
Given that ??32 \sqrt{2} = 2^{a}??, find the value of ??a??. [3]
Hide Question
Show Solution
Worked Solution
2.
Start by noting that the question is asking you to put the left hand side into a form that is expressed as a power of ??2??.
We can start by writing each term on the left hand side as a power of ??2??, starting with ??32??:
$$32$$
$$= 2^{5}$$
And then noting that Law 6 states that:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
Which allows us to rewrite ??\sqrt{2}??:
$$\sqrt{2}$$
$$=2^{\frac{1}{2}}$$
Remembering that Law 1 is:
$$x^{m} \times x^{n} = x^{(m+n)}$$
Allows us to simplify the expression given on the left hand side:
$$32 \sqrt{2}$$
$$= 2^{5} \times 2^{\frac{1}{2}}$$
$$ = 2^{5 + \frac{1}{2}}$$
$$ = 2^{\frac{11}{2}}$$
And so to answer the question:
$$a = \frac{11}{2}$$
Edexcel  Maths  Standard  2011  January  Core 1  Question 1
Question
2. (a)
Find the value of ??16^{\frac{1}{4}}??. [2]
2. (b)
Simplify ??x(2x^{\frac{1}{4}})^{4}??. [2]
Hide Question
Show Solution
Worked Solution
2. (a)
Note that Law 6 states:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
And that Law 5 states that:
$$x^{n} = \frac{1}{x^{n}}$$
So we can say that:
$$16^{\frac{1}{4}} = \frac{1}{\sqrt[4]{16}} = \frac{1}{2}$$
2. (b)
Note that Law 3 states:
$$(x^{m})^{n} = x^{mn}$$
And so we can write, noting that the power of ??\frac{3}{4}?? applies to both terms (i.e. the ??2?? and ??x^{frac{1}{4}}?? within the brackets):
$$x(2x^{\frac{1}{4}})^{4}$$
$$=x(2^{4} \times x^{\frac{1}{4} \times 4})$$
$$=x(16 \times x^{1})$$
Now we can use Law 1:
$$x^{m} \times x^{n} = x^{(m+n)}$$
Which allows us to simplify the expression we have gotten to:
$$x(16 \times x^{1})$$
$$= 16 \times x \times x^{1}$$
$$= 16 \times x^{11}$$
$$= 16 \times x^{0}$$
And finally we can use Law 4 which states that:
$$x^{0} = 1$$
To get to a final answer:
$$16 \times x^{0}$$
$$= 16 \times 1$$
$$ = 16$$
Edexcel  Maths  Standard  2011  June  Core 1  Question 1
Question
1.
Find the value of:
1. (a)
??25^{\frac{1}{2}}??. [1]
1. (b)
??25^{\frac{3}{2}}??. [2]
Hide Question
Show Solution
Worked Solution
1. (a)
Start by noting that Law 6 states that:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
So we can say that:
$$25^{\frac{1}{2}} = \sqrt{25} = 5$$
1. (b)
We know, from 1. (a) that ??25^{\frac{1}{2}} = 5?? and we also know from Law 6 that:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
And from Law 5 that:
$$x^{n} = \frac{1}{x^{n}}$$
Which means that:
$$25^{\frac{3}{2}} = \frac{1}{(\sqrt{25})^{3}} = \frac{1}{5^{3}} = \frac{1}{125}$$
Edexcel  Maths  Standard  2012  June  Core 1  Question 2
Question
2. (a)
Evaluate ??(32)^{\frac{3}{5}}??, giving your answer as an integer. [2]
2. (b)
Simplify fully ??\left(\frac{25x^{4}}{4} \right)^{\frac{1}{2}}??. [2]
Hide Question
Show Solution
Worked Solution
2. (a)
Note that Law 6 states:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
So we can say that:
$$32^{\frac{3}{5}} = (\sqrt[5]{32})^{3} = (2)^{3} = 8$$
2. (b)
Note that Law 3 states:
$$(x^{m})^{n} = x^{mn}$$
And so we can write, noting that the power of ??\frac{1}{2}?? applies to both the numerator and denominator within the brackets:
$$\left(\frac{25x^{4}}{4} \right)^{\frac{1}{2}}$$
$$=\frac{(25x^{4})^{\frac{1}{2}}}{4^{\frac{1}{2}}}$$
Note that on the numerator, the power of ??\frac{1}{2}?? applies to both terms (i.e. the ??25?? and ??x^{4}??) and we can write then:
$$\frac{(25x^{4})^{\frac{1}{2}}}{4^{\frac{1}{2}}}$$
$$=\frac{(25)^{\frac{1}{2}} \times (x^{4})^{\frac{1}{2}}}{4^{\frac{1}{2}}}$$
Note that Law 5 states that:
$$x^{n} = \frac{1}{x^{n}}$$
Which allows us to write:
$$\frac{(25)^{\frac{1}{2}} \times (x^{4})^{\frac{1}{2}}}{4^{\frac{1}{2}}}$$
$$=\frac{\frac{1}{\sqrt{25}} \times \frac{1}{\sqrt{x^{4}}} }{\frac{1}{\sqrt{4}}}$$
$$=\frac{\frac{1}{5} \times \frac{1}{x^{2}}}{\frac{1}{2}}$$
$$=\frac{\frac{1}{5x^{2}}}{\frac{1}{2}}$$
$$=\frac{2}{5x^{2}}$$
Edexcel  Maths  Standard  2013  January  Core 1  Question 2
Question
2.
Express ??8^{2x+3}?? in the form ??2^{y}??, stating ??y?? in terms of ??x??. [2]
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Show Solution
Worked Solution
2.
Start by noting that we want to write the left hand side as an expression that has ??2?? as the base, so lets split up the left hand side and write each
part with a base of ??2??. Start by using Law 1 that states:
$$x^{m} \times x^{n} = x^{(m+n)}$$
And allows us to write:
$$8^{2x+3}$$
$$= 8^{2x} \times 8^{3}$$
Now we know that ??8 = 2^{3}??, so we can rewrite this as:
$$ 8^{2x} \times 8^{3}$$
$$=(2^{3})^{2x} \times 2^{3}$$
We can next use Law 3 which states that:
$$(x^{m})^{n} = x^{mn}$$
And this allows us to write:
$$(2^{3})^{2x} \times (2^{3})^{3}$$
$$= 2^{3 \times 2x} \times 2^{3 \times 3}$$
$$= 2^{6x} \times 2^{9}$$
And finally, we can use Law 1 again:
$$ 2^{6x} \times 2^{9}$$
$$= 2^{6x + 9}$$
Which means that we can give ??y?? in terms of ??x?? as:
$$y = 6x + 9$$
Edexcel  Maths  Standard  2013  June  Core 1  Question 3
Question
3. (a)
Find the value of ??8^{\frac{5}{3}}??. [2]
3. (b)
Simplify fully ??\left( \frac{ (2x^{\frac{1}{2}})^{3} } {4x^{2}} \right)??. [3]
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Show Solution
Worked Solution
3. (a)
Note that Law 6 states:
$$x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$$
So we can say that:
$$8^{\frac{5}{3}} = (\sqrt[3]{8})^{5} = (2)^{5} = 32$$
3. (b)
Note that Law 3 states:
$$(x^{m})^{n} = x^{mn}$$
And so we can write, noting that the power of ??3?? applies to both terms within brackets (i.e. ??2?? and ??x^{\frac{1}{2}}??):
$$\left( \frac{ (2x^{\frac{1}{2}})^{3} } {4x^{2}} \right)$$
$$=\left( \frac{ 2^{3} \times (x^{\frac{1}{2}})^{3} } {4x^{2}} \right)$$
$$=\left( \frac{ 8 \times x^{\frac{3}{2}} } {4x^{2}} \right)$$
$$=\left( \frac{ 8x^{\frac{3}{2}} } {4x^{2}} \right)$$
Next we can use Law 2, which states that:
$$x^{m} \div x^{n} = x^{(mn)}$$
Which allows us to write:
$$\left( \frac{ 8x^{\frac{3}{2}} } {4x^{2}} \right)$$
$$=2 \left( \frac{ x^{\frac{3}{2}} } {x^{2}} \right)$$
$$=2 \times x^{\frac{3}{2}  2} $$
$$=2x^{\frac{1}{2}}$$
Finally we can use Law 5 which states that:
$$x^{n} = \frac{1}{x^{n}}$$
Which allows us to get to the fully simplified answer:
$$2x^{\frac{1}{2}}$$
$$=\frac{2}{x^{\frac{1}{2}}}$$
Edexcel  Maths  International  2013  June  Core 1  Question 5
Question
5.
Solve:
5. (a)
??2^{y}=8??. [1]
5. (b)
??2^{x} \times 4^{x+1} = 8??. [4]
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Show Solution
Worked Solution
5. (a)
Note that you should recognise ??2^{3} = 8?? and so the answer can just be writtedn down:
$$y = 3$$
5. (b)
Note that ??2??, ??4?? and ??8??, the bases for each term all have a common factor of ??2??. We can therefore rewrite the equation such that
only a single base, ??2??, is used:
$$2^{x} \times 4^{x+1} = 8$$
$$\implies 2^{x} \times (2^{2})^{x+1} = 2^{3}$$
Note that Law 3 states:
$$(x^{m})^{n} = x^{mn}$$
And so we can further simplify the equation:
$$2^{x} \times (2^{2})^{x+1} = 2^{3}$$
$$\implies 2^{x} \times 2^{2 \times (x+1)} = 2^{3}$$
$$\implies 2^{x} \times 2^{2x + 2} = 2^{3}$$
Noting that Law 1 states:
$$x^{m} \times x^{n} = x^{(m+n)}$$
Means that we can write:
$$ 2^{x} \times 2^{2x + 2} = 2^{3}$$
$$\implies 2^{x + 2x + 2} = 2^{3}$$
$$\implies 2^{3x + 2} = 2^{3}$$
We can now solve for ??x??:
$$2^{3x + 2} = 2^{3}$$
$$\implies 3x + 2 = 3$$
$$\implies 3x = 1$$
$$\implies x = \frac{1}{3}$$
Exam Tips

Exam questions are often simple applications of the above laws, for example, often taking the form of evaluating expressions such as:
??25^{\frac{3}{2}}??
??100^{\frac{1}{2}}??

Sometimes exam questions have focused on rewriting an expression in another form, for example:
Rewrite ??8^{3}?? as ??2^{x}??
Rewrite ??16^{\frac{3}{2}}?? as ??4^{x}??

Remember, there is no calculator allowed, so the examiners won’t ask you to complete complicated arithmetic.Consider what order you do any calculation in to make the arithmetic easier. For the questions below, the easy way is shown in green and the hard way in red.
Find ??25^{\frac{3}{2}}??.
??25^{\frac{3}{2}} = \sqrt{25 \times 25 \times 25} = \sqrt{15625} = 125??.
??25^{\frac{3}{2}} = \sqrt{25} \times \sqrt{25} \times \sqrt{25} = 5 \times 5 \times 5 = 125??.
Find ??27^{\frac{4}{3}}??.
??27^{\frac{4}{3}} = \sqrt[3]{27^{4}} = \sqrt[3]{27 \times 27 \times 27 \times 27} = =\sqrt[3]{531441} = 81??.
??27^{\frac{4}{3}} = \sqrt[3]{27} \times \sqrt[3]{27} \times \sqrt[3]{27} \times \sqrt[3]{27} = 3 \times 3 \times 3 \times 3 = 81??.

You should learn and become familiar with the laws above. None of the information from this page is in the materials given to you in the exam ('Mathematical Formulae') and the methods used here will often be
required in other questions, often, for example, in questions about surds.