# Negative Powers

Here we will learn about negative powers, including what they are, how to simplify and evaluate them, and how to combine them with other laws of exponents.

There are also powers and roots worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What are negative powers?

Negative powers are a type of index that represents a number that can be written as a fraction.

For example,

3^{-2}=\frac{1}{3^{2}}=\frac{1}{9} .

A negative power does not mean that the base is a negative number.
If the base is a positive number, the denominator of the fraction remains positive.

Let’s see why a negative power is a fraction using the powers of 10 .

As we divide each row by 10 , the power is reduced by 1 .

When we reach 100 (the zero exponent) we are dividing the base number by itself, any base to the power of 0 is always equal to 1 .

When we divide by a further 10 , we get  0.1=\frac{1}{10}=10^{-1} .

As we continue to divide by 10 , the denominator of the fraction becomes increasingly large, meaning that the original number gets smaller and smaller (closer to 0 ).

### What are negative powers? ## Positive reciprocal

Another term for the negative power is the positive reciprocal.

The reciprocal of a number is the value that can be multiplied by the original number to get the answer of 1 .

For example,

2\times\frac{1}{2}=1 and so \frac{1}{2} is the positive reciprocal of 2 .

Introducing a negative power, we get 2^{-1}=\frac{1}{2^{1}}=\frac{1}{2} which we can express in a general form as

x^{-a}=\frac{1}{x^{a}}

where x and a are constants.

Most of the laws of indices can include a negative power and the toughest questions usually involve a fractional base number raised to a negative fraction.

For example,

(\frac{25}{4})^{-\frac{3}{2}}=\frac{8}{125} .

## Simplify and evaluate

Questions will often require you to simplify or evaluate an expression.

These terms are subtly different and so the table below shows a couple of examples of the difference between them.

Simplify (2^{2})^{-3} Evaluate (2^{2})^{-3}
Simplify: to make something easier to understand.Evaluate: to find the value of an expression.
(2^{2})^{-3}=2^{2\times -3}=2^{-6} …=\frac{1}{2^{6}}=\frac{1}{64}
So (2^{2})^{-3}=2^{-6} as 2^{-6} is more simple to understand…… but the value of the expression (2^{2})^{-3}=\frac{1}{64} .

Remember: A fraction can also be converted to a decimal.

## How to operate with negative powers

In order to use negative powers:

1. Simplify any powers using laws of indices.
2. Evaluate or solve (if required).

### How to operate with negative powers ### Related lessons onpowers and roots

Negative powers is part of our series of lessons to support revision on powers and roots. You may find it helpful to start with the main powers and roots lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

## Negative powers examples

### Example 1: simplify and evaluate

Simplify and evaluate the expression.

6^{-2}\div 6^{-4}

1. Simplify any powers using laws of indices.

As x^{a}\div x^{b}=x^{a-b} , we have 6^{-2}\div 6^{-4} = 6^{-2--4}=6^{-2+4}=6^{2} .

2Evaluate or solve (if required).

Evaluating the expression 6^{2} , we get 6^{2}=6\times 6=36.

So 6^{-2}\div 6^{-4}=36 .

### Example 2: algebraic terms

Write the following expression as a fraction in its simplest form.

2y^{-a}\times 2x^{b}

Simplify any powers using laws of indices.

Evaluate or solve (if required).

### Example 3: inequalities

Solve the inequality.

(3x)^{-2}<\frac{1}{36}

Simplify any powers using laws of indices.

Evaluate or solve (if required).

### Example 4: negative fractional power

Simplify and evaluate the following expression.

125^{-\frac{2}{3}}

Simplify any powers using laws of indices.

Evaluate or solve (if required).

### Example 5: fractional base number

Simplify fully, (\frac{3}{4})^{-3} .

Simplify any powers using laws of indices.

Evaluate or solve (if required).

### Example 6: expressing a power with a different base

Simplify fully, (4^{-10})^{\frac{1}{2}} .

Simplify any powers using laws of indices.

Evaluate or solve (if required).

### Common misconceptions

• Negative powers and negative number

A common error is to think that a negative power suggests that the entire number is a negative number. This is incorrect because a negative power means that we are finding a positive reciprocal of the base.

For example,

2^{-4}=-2^{4}=-16 is incorrect.

The correct answer is 2^{-4}=\frac{1}{2^{4}}=\frac{1}{16} .

• Changing the base incorrectly

For example,

5^{3}\times2^{-4}=(5\times{2})^{3-4}=10^{-1}=\frac{1}{10} .

This is incorrect as the bases must be the same when simplifying calculations with powers.

The correct answer would be 5^{3}\times{2^{-4}}=5^{3}\times{\frac{1}{16}}=\frac{125}{16} .

• Subtracting a negative number

A common error is to incorrectly subtract negative numbers.

For example,

3^{-2}\div{3^{-5}}=3^{-2-5}=3^{-7}=\frac{1}{3^{7}}

This is incorrect because the 5 is a negative number, and subtracting a negative number means that we add.

The correct answer would be 3^{-2}\div 3^{-5}=3^{-2--5}=3^{-2+5}=3^{3}=27.

• BIDMAS

When raising a base to a power, the power is associated with a variable or an expression.

For example, let’s look at  3x^{-2}.

Using BIDMAS we can see that, x is being raised to the power of -2 , and then is multiplied by 3 to get the answer \frac{3}{x^{2}} .

However a common error is to calculate 3x raised to the power of -2 , giving the answer \frac{1}{9x^{2}} which is incorrect. If this was the case, the question would be written as (3x)^{-2}.

• Incorrect application of the exponent

A similar circumstance is applied to fractions that are raised to a power.

For example, let’s look at (\frac{5}{2})^{-3} .

The correct application of the laws of indices would give a correct answer of

(\frac{5}{2})^{-3}=(\frac{2}{5})^{3}=\frac{8}{125} .

However a common error would be to apply the -3 only to the numerator, leaving the denominator unchanged. This would give the incorrect answer of

(\frac{5}{2})^{-3}=\frac{1}{125}\times\frac{1}{2}=\frac{1}{250} .

### Practice negative powers questions

1. Evaluate 5^{4}\times 5^{-6} .

25 -25 -\frac{1}{25} \frac{1}{25} 5^{4}\times{5^{-6}}=5^{4-6}=5^{-2}=\frac{1}{5^{2}}=\frac{1}{25}

2. Simplify 3x^{a}\div 6^{y-b} fully.

\frac{x^{a}}{y^{b}} \frac{x^{a}}{2y^{b}} \frac{x^{a}y^{b}}{2} \frac{2}{x^{a}y^{b}} 3x^{a}\div{6y^{-b}}=3x^{a}\div{\frac{6}{y^{b}}}

=3x^{a}\times{\frac{y^{b}}{6}}=\frac{3x^{a}y^{b}}{6}=\frac{x^{a}y^{b}}{2}

3. State the range of values for the following inequality, x^{-2}>\frac{1}{16} .

-4<x<4 x<4 -\frac{1}{4}<x<\frac{1}{4} x>4, x<-4   4. Evaluate 16^{\frac{-3}{4}} .

\frac{1}{64} -12 \frac{1}{8} -64  5. Evaluate (\frac{1}{5})^{-2} .

\frac{1}{25} \frac{-2}{5} 25 -\frac{1}{5} (\frac{1}{5})^{-2}=(\frac{5}{1})^{2}=5^{2}=25

6. Simplify the expression below. Write your answer in the form 2^{n} .

(8^{2})^{-\frac{1}{3}}

2^{2} 2^{-2} 2^{-1} 2^{\frac{-2}{3}} (8^{2})^{-\frac{1}{3}}=(2^{2\times{3}})^{-\frac{1}{3}}=(2^{6})^{-\frac{1}{3}}=2^{-\frac{6}{3}}=2^{-2}

### Negative powers GCSE questions

1. Order the numbers smallest to largest.

6\times 10^{-4} ,   0.006\times 10^{5} ,   6\times 10^{0} ,   10^{-4}\times 2^{3} ,   0.06\times 10^{1}

(3 marks)

6\times 10^{-4} = 0.0006\\ 10^{-4}\times 2^{3} = 0.0008\\ 0.06\times 10^{1}=0.6\\ 6\times 10^{0} = 6\\ 0.006\times 10^{5}= 600\\

(1)

(2)

(3)

2. (a) Solve x^{-3}=\frac{1}{125} .

(b) Solve the inequality x^{-2}>16 .

(6 marks)

(a)

x^{-3}=\frac{1}{x^{3}}

(1)

x^{3}=125

(1)

x=5

(1)

(b)

x^{-2}>\frac{1}{x^{2}}

(1)

x^{2}>\frac{1}{16}

(1)

x<-\frac{1}{4} or x>\frac{1}{4}

(1)

3. Use laws of indices to simplify and evaluate

\frac{7^{5}\times{7^{-2}}}{7^{6}} .

(3 marks)

7^{5}\times 7^{-2}=7^{5+-2}=7^{3}

(1)

7^{3}\div 7^{6}=7^{3-6}=7^{-3}

(1)

7^{-3}=\frac{1}{7^{3}}=\frac{1}{343}

(1)

## Learning checklist

You have now learned how to:

• Extend your understanding of the number system to include powers
• Calculate with roots, and with integer indices
• Simplify expressions involving sums, products and powers, including the laws of indices

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