Indices

UK Secondary (7-11)

Power of a power with integer bases and negative powers

Lesson

We've already looked at the power of a power rule, which states:

$\left(a^m\right)^n=a^{mn}$(`a``m`)`n`=`a``m``n`

Now we are going to use apply this rule to terms involving negative bases, and also negative indices.

Remember!

The base number stays the same. We just multiply the powers.

When we raise a negative numbers by an even power, we get a positive answer e.g. $\left(-4\right)^2=-4\times\left(-4\right)$(−4)2=−4×(−4)$=$=$16$16

The same pattern applies to negative powers: $\left(-2\right)^{-4}=2^{-4}$(−2)−4=2−4$=$=$\frac{1}{16}$116

This means we can simplify expressions involving negative numbers with negative indices, and express them as a fraction with a positive index. If the power is even the fraction will be positive, but if the power is odd the fraction will be negative.

For example, $\left(-5\right)^{-3}=\frac{1}{\left(-5\right)^3}$(−5)−3=1(−5)3 $=$= $-\frac{1}{125}$−1125

Simplify the following into the form $a^b$`a``b`:

$\left(6^7\right)^{-3}$(67)−3

Evaluate $\left(5^{-9}\right)^0$(5−9)0.

$\left(\left(-4\right)^{-8}\right)^2$((−4)−8)2 simplifies to which of the following:

$-4^{-6}$−4−6

A$4^{-6}$4−6

B$4^{-16}$4−16

C$-4^{-16}$−4−16

D$-4^{-6}$−4−6

A$4^{-6}$4−6

B$4^{-16}$4−16

C$-4^{-16}$−4−16

D

We want to simplify the following expression using index laws: $4^{-2}\times64^{-3}$4−2×64−3.

To use the index laws, we need both terms to be written with the same base.

Fill in the box to re-write $64^{-3}$64−3 with a base of $4$4.

$64^{-3}=\left(4^{\editable{}}\right)^{-3}$64−3=(4)−3

Using the result of the previous part, express $4^{-2}\times64^{-3}$4−2×64−3 in simplest positive index form.