Hong Kong
Stage 1 - Stage 3

# Power of a power with variable bases

Lesson

In Power of powers, we learnt the process of how to simplify a term expressed as the power of a power.

The power of a power law

For any base number $a$a, and any numbers $m$m and $n$n as powers,

$\left(a^m\right)^n=a^{m\times n}$(am)n=am×n

That is, when simplifying a term with a power that itself has a power:

• Keep the same base
• Find the product of the powers

Here we will look at powers that have variable bases, and also a mix of variable and integer bases. Since "powers of powers" involve expressions with brackets, it's important to remember that everything inside the brackets is raised to the outside power.

#### Exploration

Let's say we want to simplify the expression $\left(2x^2\right)^3$(2x2)3:

A common mistake is to only apply the outside power to the algebraic term. If we did this, we would get an answer of $2x^{2\times3}=2x^6$2x2×3=2x6, which is not correct.

Consider the expression in expanded form: $\left(2x^2\right)^3=2x^2\times2x^2\times2x^2$(2x2)3=2x2×2x2×2x2

 $\left(2x^2\right)^3$(2x2)3 $=$= $2x^2\times2x^2\times2x^2$2x2×2x2×2x2 $=$= $\left(2\times2\times2\right)\times\left(x^2\times x^2\times x^2\right)$(2×2×2)×(x2×x2×x2) $=$= $2^3\times\left(x^2\right)^3$23×(x2)3 $=$= $8x^6$8x6

You can see that not only is $x^2$x2 multiplied $3$3 times, $\left(x^2\right)^3$(x2)3, but $2$2 is also multiplied $3$3 times, $2^3$23.

So we need to raise $2$2 to the power of $3$3 as well as $x^2$x2 to the power of $3$3.

$\left(2x^2\right)^3=8x^6$(2x2)3=8x6

Careful!
Beware of the signs when raising negative numbers to a power, as a negative number raised to an odd power has a negative result. For example:

$\left(-2x^2\right)^3=-8x^6$(2x2)3=8x6

#### Examples

##### Question 1

Simplify the following, giving your answer with a positive index: $\left(w^3\right)^4$(w3)4

##### Question 2

Simplify, and evaluate where possible, the following expression:

$\left(-2x^2\right)^2$(2x2)2

##### Question 3

Simplify the following expression: $\left(3a^3b^2c^3\right)^3$(3a3b2c3)3.