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Stage 1 - Stage 3

# Index notation with variable bases and negative powers

Lesson

We have looked before at how to manipulate negative indices.

Remember

The negative index law states:

$a^{-x}=\frac{1}{a^x}$ax=1ax

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.

We now need to consider what happens if we have a coefficient as well as an unknown raised to a negative power.

#### Examples

##### Question 1

Express $3x^{-3}$3x3 with a positive index.

Think - the $x$x has been raised to a power of $-3$3. We must evaluate this first, then multiply by $3$3

Do - Apply the negative index law, then multiply by $3$3.

 $3x^{-3}$3x−3 $=$= $3\times x^{-3}$3×x−3 $=$= $3\times\frac{1}{x^3}$3×1x3​ $=$= $\frac{3}{x^3}$3x3​

As the power is only on the $x$x, the value of the coefficient hasn't changed.

## A Fraction to a Negative Power

Things become a bit more complicated when we need to raise a fraction to a negative power.

First, you may want to remind yourself what we mean by a reciprocal.

The reciprocal of $\frac{1}{8}$18 is $8$8. The reciprocal of $\frac{2}{3}$23 is $\frac{3}{2}$32. The reciprocal of $91$91 is $\frac{1}{91}$191.

Can you see what reciprocal means? Simply, if you have a fraction, you need to invert it (or flip it) to find the reciprocal. If you have an integer, you put that integer as the denominator, and $1$1 as the numerator.

This links to negative powers, as we saw above.

#### Example

##### Question 2

So what happens if you are asked to calculate $\left(\frac{a}{b}\right)^{-3}$(ab)3?

 $\left(\frac{a}{b}\right)^{-3}$(ab​)−3 $=$= $\frac{a^{1\times\left(-3\right)}}{b^{1\times\left(-3\right)}}$a1×(−3)b1×(−3)​ First we need to multiply out the powers. $=$= $\frac{a^{-3}}{b^{-3}}$a−3b−3​ We know that this is the same as $=$= $a^{-3}\div b^{-3}$a−3÷​b−3 Now let's rewrite this expression using only positive powers $=$= $\frac{1}{a^3}\div\frac{1}{b^3}$1a3​÷​1b3​ Dividing by a fraction is the same as multiplying by the reciprocal of that fraction $=$= $\frac{1}{a^3}\times b^3$1a3​×b3 Can you see what is going to happen next? By multiplying through we now get $=$= $\frac{b^3}{a^3}$b3a3​ Let's take a moment to compare this answer to the question.

$\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}$(ab)3=b3a3

Look for a shortcut here.

What has happened is we have found the reciprocal of the question, and raised each term to the power which is now a positive.

##### Question 3

Now let's try another.

Express $\left(\frac{x^8}{y^5}\right)^{-4}$(x8y5)4  with a positive index.

 $=$= $\frac{x^{8\times\left(-4\right)}}{y^{5\times\left(-4\right)}}$x8×(−4)y5×(−4)​ $=$= $\frac{x^{-32}}{y^{-20}}$x−32y−20​ $=$= $\frac{y^{20}}{x^{32}}$y20x32​

Now it's your turn to practice applying these rules.

#### Worked Examples

##### Question 4

Express $4y^{-2}\times2y^{-4}$4y2×2y4 with a positive index.

##### Question 5

Express $\left(\frac{x^2}{y^4}\right)^{-1}$(x2y4)1 without negative indices.