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Index notation with variable bases and negative powers


We have looked before at how to manipulate negative indices. 


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.

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



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}$3x3 $=$= $3\times x^{-3}$3×x3
  $=$= $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.



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.
  $=$= $\frac{a^{-3}}{b^{-3}}$a3b3 We know that this is the same as
  $=$= $a^{-3}\div b^{-3}$a3÷​b3 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. 


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)
  $=$= $\frac{x^{-32}}{y^{-20}}$x32y20
  $=$= $\frac{y^{20}}{x^{32}}$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.


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