Dealing with nasty powers

There’s nearly always a question on the non-calculator GCSE paper about Nasty Powers. I’m not talking about the Evil Empire or anything, I just mean powers that aren’t nice - we can all deal with positive integer powers, it’s the zeros, the negatives and the fractions that get us down.

Zeroth powers

What’s ${10}^3$? It’s 1,000, clearly. Everyone knows that.

${10}^2$? A hundred. These are getting easier!

${10}^1$? Well… it’s ten, isn’t it? There’s a pattern. We’re dividing by ten each time.

So, that would make… ${10}^0=1$. (That makes sense with the power laws, as well: ${10}^2÷{10}^2={10}^0$, and $100÷100=1$, so ${10}^0=1$.)

In fact, anything ¹ to the power of 0, is 1. ${12}^0=1$. $(-4{)}^0=1$. ${2014}^0=1.$ $x^0=1$, as long as $x$ isn’t zero.

Negative powers

The pattern continues, though: if you know every time you knock one off the power, you divide by the base, you see straight away that ${10}^{-1}=1÷10=0.1$ (or $\frac{1}{10}$, if you’re a grown-up). ${10}^{-2}=0.01=\frac{1}{100}$.

Again, you can show this using the power laws: ${10}^{-2}={10}^{(0-2)}={10}^0÷{10}^2=1÷100$. Simple!

In general, $x^{-n}=1÷x^n$ - again, as long as $x$ isn’t zero ²

Fractional powers

What happens when you – for example – halve a power? Let’s start with $2^6=64$. Halving the power takes you to $2^3=8$. How are they linked?

Not sure? Try $3^4=81$ and $3^2=9$, or $5^2=25$ and $5^1=5$.

Halving a power gives you the square root of the number. Similarly, dividing a power by three gives you a cube root. Try it with a few!

In particular, $x^{\frac{1}{2}}$ is the same thing as $\sqrt{x}$, and $x^{\frac{a}{b}}=\sqrt[b]{x^a}$. The bottom of the power is the root you take, and the top of the fraction is the power you take.

A recipe (finally)

So, how would you deal with something like ${81}^{-\frac{3}{4}}$?

The thing to do is take it step by step – and I’d do the hardest steps first. The nastiest thing is the four on the bottom of the fraction – that means, ‘take a fourth root’. Luckily, we know that the fourth root of 81 is 3. (If we don’t know that, we can say the square root is 9, so the fourth root is the square root of 9, which is 3). The expression is already looking nicer: $3^{-3}$.

The next nasty thing is the negative power, but that just means we need to turn it into a fraction: $\frac{1}{3^3}$.

Lastly, we know that $3^3=27$, so we end up with $\frac{1}{27}$.

Edited April 19th to fix typo.

Footnotes:

1. except 0

2. Dividing by zero makes bad things happen. Don’t try it.