Exponent notation with integer and variable bases and negative powers

We've already learnt the division law which states: $a^x\div a^y=a^{x-y}$ax÷​ay=ax−y.

But what happens when $y$y is bigger than $x$x? For example, if I simplified $a^3\div a^5$a3÷​a5 using the division law, I would get $a^{-2}$a−2. So what does a negative exponent mean? Let's expand the example to find out:

Remember that when we are simplifying fractions, we are looking to cancel out common factors in the numerator and denominator. Remember that any number divided by itself is $1$1.

So using the second approach, we can also express $a^3\div a^5$a3÷​a5 with a positive exponent as $\frac{1}{a^2}$1a2​.

 

The Negative Exponent Law

We saw that $a^{-2}$a−2 is equivalent to $\frac{1}{a^2}$1a2​.

More generally:

$a^{-x}=\frac{1}{a^x}$a−x=1ax​

where $x$x is not equal to $0$0.

In other words, $a^{-x}$a−x is the reciprocal (i.e. the flipped version) of $a^x$ax. So $a^{-x}$a−x actually represents a fraction, NOT a negative number.

 

Common Fractions

Consider the number represented by $2^{-3}$2−3. What is this number?

Well we can rewrite it in positive exponential form by taking the reciprocal: $2^{-3}=\frac{1}{2^3}$2−3=123​

But $\frac{1}{2^3}=\frac{1}{8}$123​=18​. So we have found another way to represent the fraction $\frac{1}{8}$18​: $2^{-3}=\frac{1}{8}$2−3=18​

There are so many other fractions that can be represented in negative exponential form. What others can you think of?

 

Fractions raised to Negative Powers

What do we make of an expression such as $\left(\frac{5}{2}\right)^{-2}$(52​)−2 or $\frac{4^{-3}}{5^{-1}}$4−35−1​, where the numerator and denominator are both raised to negative powers?

Let's take it slowly through this $\left(\frac{5}{2}\right)^{-2}$(52​)−2and see if we can come up with a shortcut.

First of all, $\left(\frac{5}{2}\right)^{-2}$(52​)−2 can be split up in a couple of different ways. I am going to rewrite it as $\frac{5^{-2}}{2^{-2}}=5^{-2}\div2^{-2}$5−22−2​=5−2÷​2−2.

Now let's look at each part separately: $5^{-2}=\frac{1}{5^2}$5−2=152​                      $2^{-2}=\frac{1}{2^2}$2−2=122​

Of course these fractions can be simplified but let's keep them in exponential form for now:

Let's see how far we've come from the original expression:

$\left(\frac{5}{2}\right)^{-2}$(52​)−2	$=$=	$5^{-2}\div2^{-2}$5−2÷​2−2
	$=$=	$\frac{1}{5^2}\div\frac{1}{2^2}$152​÷​122​	(since dividing by a fraction is multiplying by its reciprocal)
	$=$=	$\frac{1}{5^2}\times\frac{2^2}{1}$152​×221​
	$=$=	$\frac{2^2}{5^2}$2252​
	$=$=	$\left(\frac{2}{5}\right)^2$(25​)2

Compare $\left(\frac{2}{5}\right)^2$(25​)2 and $\left(\frac{5}{2}\right)^{-2}$(52​)−2. It seems that we can make the powers positive by switching the numerator and denominator around.

 

Examples

Question 1

Simplify $\left(\frac{c}{d}\right)^{-4}$(cd​)−4

Solution:

$\left(\frac{c}{d}\right)^{-4}=\left(\frac{d}{c}\right)^4$(cd​)−4=(dc​)4

 

Question 2

Question 3

Question 4