topic badge

3.04 Negative exponents

Lesson

 

The negative power property

The negative power property states:

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

So if you need to express a negative exponent as a positive exponent, or a positive exponent as a negative exponent, you need to convert it to a fraction using this rule.

 

Worked example

Question 1

Express $3x^{-3}$3x3 with a positive exponent.

Think - the $x$x has been raised to a power of $-3$3. We must evaluate this first, then multiply by $3$3

Do - Apply the negative power property, then multiply by $3$3.

$3x^{-3}$3x3 $=$= $3\times x^{-3}$3×x3
  $=$= $3\times\frac{1}{x^3}$3×1x3
  $=$= $\frac{3}{x^3}$3x3

As the power is only on the $x$x, the value of the coefficient hasn't changed.

 

Raise a fraction to a negative power

Things become a bit more complicated when we need to raise a fraction to a negative power.

First, you may want to remind yourself what we mean by a reciprocal.

The reciprocal of $\frac{1}{8}$18 is $8$8. The reciprocal of $\frac{2}{3}$23 is $\frac{3}{2}$32. The reciprocal of $91$91 is $\frac{1}{91}$191.

Can you see what reciprocal means? Simply, if you have a fraction, you need to invert it (or flip it) to find the reciprocal. If you have an integer, you put that integer as the denominator, and $1$1 as the numerator.

This links to negative powers, as we saw above.

 

Worked example

Question 2

Calculate $\left(\frac{a}{b}\right)^{-3}$(ab)3

Think:

$\left(\frac{a}{b}\right)^{-3}$(ab)3 $=$= $\frac{a^{1\times\left(-3\right)}}{b^{1\times\left(-3\right)}}$a1×(3)b1×(3) First we need to multiply out the powers.
  $=$= $\frac{a^{-3}}{b^{-3}}$a3b3 We know that this is the same as
  $=$= $a^{-3}\div b^{-3}$a3÷​b3 Now let's rewrite this expression using only positive powers
  $=$= $\frac{1}{a^3}\div\frac{1}{b^3}$1a3÷​1b3 Dividing by a fraction is the same as multiplying by the reciprocal of that fraction
  $=$= $\frac{1}{a^3}\times b^3$1a3×b3 Can you see what is going to happen next? By multiplying through we now get
  $=$= $\frac{b^3}{a^3}$b3a3 Let's take a moment to compare this answer to the question. 

 

Solve: $\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}$(ab)3=b3a3 

Reflect: Can you see a shortcut here?

What has happened is we have found the reciprocal of the question, and raised each term to the power which is now a positive. 

 

Multiplication and division properties with negative exponents

We've already learned about the properties for exponents with multiplication and division, as well as the negative power property. Now we are going to combine these rules to simplify expressions which involve multiplication or division, and negative exponents.

Consider the expression: $e^7\times e^{-4}$e7×e4

Notice the following: 

  • There is multiplication and the bases are the same (we can apply the multiplication law)
  • One of the powers is negative (we can express the second term with a positive power if we wish)

When negative powers are involved, this opens up choices in how we go about trying to simplify the expression.

With the above example, I have two choices:

One Approach: Add the powers immediately as the bases are the same and we are multiplying 

$e^7\times e^{-4}$e7×e4 $=$= $e^{7+\left(-4\right)}$e7+(4)  
  $=$= $e^{7-4}$e74 (recall that a plus and minus sign next to each other result in a minus)
  $=$= $e^3$e3  

 

Another Approach: First express the second term with a positive power

$e^7\times e^{-4}$e7×e4 $=$= $e^7\times\frac{1}{e^4}$e7×1e4  
  $=$= $\frac{e^7}{e^4}$e7e4  
  $=$= $e^{7-4}$e74 (subtract the powers using the division rule)
  $=$= $e^3$e3  

Of course, which way you go about these types of problems is completely up to you.

 

Worked example

Question 3

Simplify the following, giving your answer in exponential form: $\frac{35x^2y^2}{-5x^{14}y^{14}}$35x2y25x14y14.

Think: Let's separate the terms and apply the Exponent Division law.

Do

$35\div\left(-5\right)$35÷​(5) $=$= $-7$7 the coefficient term
$x^2\div x^{14}$x2÷​x14 $=$= $x^{2-14}$x214 the $x$x's
  $=$= $x^{-12}$x12 the $x$x's
  $=$= $\frac{1}{x^{12}}$1x12 and take the reciprocal
$y^2\div y^{14}$y2÷​y14 $=$= $y^{2-14}$y214 the $y$y's
$y^2\div y^{14}$y2÷​y14 $=$= $y^{-12}$y12 the $y$y's
  $=$= $\frac{1}{y^{12}}$1y12

and take the reciprocal

  $=$= $-7\times\frac{1}{x^{12}}\times\frac{1}{y^{12}}$7×1x12×1y12 all together
$\frac{35x^2y^2}{-5x^{14}y^{14}}$35x2y25x14y14 $=$= $\frac{-7}{x^{12}y^{12}}$7x12y12 and simplify

 

Power of a power rule with negative exponents

We already learned to apply the power of a power rule to positive exponents. This rule states:

$\left(x^a\right)^b=x^{a\times b}$(xa)b=xa×b

Now we are going to explore what happens when we also include negative values in these kinds of questions.

If you think back to learning about multiplying and dividing by negative numbers, you'll remember that:

  • Multiplying a positive and a negative value will give you a negative answer.
  • Multiplying two negative numbers together gives you a positive answer.

Let's think for a minute about what happens if we multiply a negative value by itself more than twice. 

Say we had the question $\left(-2\right)^3$(2)3. This means $-2\times\left(-2\right)\times\left(-2\right)$2×(2)×(2). If  we simplify this, $-2\times\left(-2\right)=4$2×(2)=4 and $-4\times\left(-2\right)=-8$4×(2)=8.

What about if we multiplied it by itself again to get the answer to $\left(-2\right)^4$(2)4? We know that $\left(-2\right)^3=-8$(2)3=8 and $-8\times\left(-2\right)=16$8×(2)=16.

So, as a general rule:

  • If you multiply a negative number by a positive power, you will get a positive answer
  • If you multiply a negative number by a negative power, you will get a negative answer.

 

Worked example

Question 4

Simplify $\left(w^{10}\right)^{-5}$(w10)5.

Think: Using the power to a power rule, we need to multiply the exponents.

$\left(w^{10}\right)^{-5}=w^{10\times\left(-5\right)}$(w10)5=w10×(5)

Do: $\left(w^{10}\right)^{-5}=w^{-50}$(w10)5=w50

Simplify the negative exponent: $\left(w^{10}\right)^{-5}=\frac{1}{w^{50}}$(w10)5=1w50

ReflectHow would this problem be different if both powers were negative?

 

Practice questions

Question 1

Simplify the following, giving your answer with a positive exponent: $\frac{9x^3}{3x^{-4}}$9x33x4

Question 2

Simplify the following, writing without negative exponents.

$5p^2q^{-4}\times8p^{-2}q^6$5p2q4×8p2q6

Question 3

Express $4y^{-2}\times2y^{-4}$4y2×2y4 with a positive exponent.

Question 4

Express $\left(\frac{x^2}{y^4}\right)^{-1}$(x2y4)1 without negative exponents.

 

Outcomes

8.NS.3

Given a numeric expression with common rational number bases and integer exponents, apply the properties of exponents to generate equivalent expressions

What is Mathspace

About Mathspace