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# Indices: Brackets and Combinations

### Indices: Brackets and Combinations

We can raise an index to another index, and combine with other rules such as multiplication and division.

Sometimes we need to raise a power to a power. For example:

$(x^3)^2$

Here, we are squaring $x^3$, what is another way of writing $(x^3)^2$?

We now have a multiplication, what is $x^3 \times x^3$?

We have now found our answer!

$(x^3)^2=x^6$

Now we know that $(x^3)^2=x^6$, how are the powers related?

We have found our rule!

When raising a power to a power, shown by brackets, multiply the powers.

We can use the rule to simplify $(4^2)^4$ as a single power of $4$.

$(4^2)^4=4^{2 \times4}=4^8$

Try this one, what is $(6^5)^3$ as a power of 6?

What is $(11^{12})^9$ as a power of $11$?

What happens if the power is 0? Like in$3^0$?

Indices follow a pattern...

Let's use powers of $3$ to identify this pattern.

What is $3^3$

We've found $3^3$, dividing this by $3$ gives $\dfrac{3^3}{3}=3^2$. What is $3^2$?

Similarly $\dfrac{3^2}{3}=3^1$. What is $3^1$?

Each time the power is reduced by $1$, we have divided by $3$.

$\dfrac{3^3}{3}=3^2$ and $\dfrac{3^2}{3}=3^1$

Continuing that pattern, what do we get in terms of powers of $3$ when we divide $\dfrac{3^1}{3}$?

Okay, so what does $3^0$ equal?

To find $3^0$, we need to calculate $3 \div 3$

What is $3\div3$?

$3^0 = 1$

This is the same for **any number** to the power of 0.

What is the value of $12^0$?

Let's combine these rules and simplify $\dfrac{(x^6)^2}{x^3 \times x^4}$

What are we left with when we simplify the numerator $(x^6)^2$?

Now simplify the denominator $x^3 \times x^4$

Our expression can be rewritten as $\dfrac{x^{12}}{x^{7}}$

By subtracting the index numbers (since it is a division), we can find the simplest version.

Simplify $\dfrac{x^{12}}{x^{7}}$

Let's try another:

$\bigg(\dfrac{x^{13}}{x^5}\bigg)^4$

Divide the powers inside the bracket

Now raise to the power of $4$

Nice! 👍

The final answer is $x^{32}$.

Summary! When raising a power to a power, shown by using brackets, multiply the powers.

$(x^3)^4=x^{12}$

Any number raised to the power $0$ is $1$.

$12^0=7^0=1$

The rules can be combined to simplify complex expressions. All the rules depend on the base number being the same.

$\dfrac{(x^7)^2}{x^7\times x^9}=\dfrac{x^{14}}{x^{16}}=x^{-2}$