UK Secondary (7-11)
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Power of a power with variable bases and negative powers

We already learnt to apply the power of a power rule to positive indices. 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 raise a negative number by an even index, you will get a positive answer
  • If you raise a negative number by an odd index, you will get a negative answer



The negative index law states:


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



Question 1

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

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


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


Question 2

Express $\left(5y^3\right)^{-3}$(5y3)3 with a positive index.

Question 3

Simplify the following:



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