Laws of exponents

Powers with whole numbers

You know that $a^3 = a \times a \times a$. For a positive number $a$ and whole numbers $m$ and $n$, there are five laws.

1. $a^m \times a^n = a^{m+n}$
2. $(a^m)^n = a^{mn}$
3. $\dfrac{a^m}{a^n} = a^{m-n}$
4. $a^m \times b^m = (ab)^m$
5. $a^0 = 1$, and $a^{-n} = \dfrac{1}{a^n}$

Examples: $2^3 \times 2^5 = 2^8 = 256$. $(2^3)^2 = 2^6 = 64$. $\dfrac{3^5}{3^2} = 3^3 = 27$. $5^{-2} = \dfrac{1}{25}$.

Why do the laws work?

The first law is just counting. $2^3 \times 2^5$ is $(2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2 \times 2)$, which is eight twos multiplied, that is $2^8$.

The third law with $m = n$ explains $a^0$. We know $\dfrac{a^3}{a^3} = 1$. By the law, it is $a^{3-3} = a^0$. So $a^0$ has to be $1$ for the law to work.

In the same way, $\dfrac{a^2}{a^5} = a^{-3}$ by the law. But we can also cancel: $\dfrac{a^2}{a^5} = \dfrac{1}{a^3}$. So $a^{-3} = \dfrac{1}{a^3}$. A negative power means a reciprocal. For example, $2^{-1} = \dfrac{1}{2}$, and $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$.

Fractional powers

What could $a^{1/2}$ mean? By the first law, $a^{1/2} \times a^{1/2} = a^1 = a$. So $a^{1/2}$ is the number that, multiplied by itself, gives $a$. It is the square root: $a^{1/2} = \sqrt{a}$.

In the same way, $a^{1/n}$ is the $n$-th root of $a$, the number whose $n$-th power is $a$.

Examples:

• $64^{1/2} = 8$, because $8^2 = 64$.
• $27^{1/3} = 3$, because $3^3 = 27$.
• $8^{2/3} = (8^{1/3})^2 = 2^2 = 4$.
• $16^{3/4} = (16^{1/4})^3 = 2^3 = 8$.
• $81^{-1/4} = \dfrac{1}{81^{1/4}} = \dfrac{1}{3}$.

Always take the root first, then the power. For $8^{2/3}$, the root $8^{1/3} = 2$ is small, and then $2^2 = 4$. If we took the power first, $8^2 = 64$, and then the cube root of $64$, we would get the same answer, with harder numbers.

Keep $a = 8$ and set $n = 3$. Move $m$ from $-2$ to $4$. The red dot steps along the curve through $\tfrac{1}{4}$, $\tfrac{1}{2}$, $1$, $2$, $4$, $8$ and $16$.

The laws hold for fractional powers

The same five laws hold for fractional powers.

• $2^{2/3} \times 2^{1/3} = 2^{2/3 + 1/3} = 2^1 = 2$.
• $\left(3^{1/2}\right)^4 = 3^{2} = 9$.
• $5^{1/2} \times 7^{1/2} = (35)^{1/2} = \sqrt{35}$.
• $\dfrac{7^{3/4}}{7^{1/4}} = 7^{1/2} = \sqrt{7}$.

A longer example

Simplify $\dfrac{9^{3/2} \times 2^{-2}}{4^{-1/2}}$.

• $9^{3/2} = (9^{1/2})^3 = 3^3 = 27$.
• $2^{-2} = \dfrac{1}{4}$.
• $4^{-1/2} = \dfrac{1}{4^{1/2}} = \dfrac{1}{2}$.

So the value is $\dfrac{27 \times \frac{1}{4}}{\frac{1}{2}} = \dfrac{27}{4} \times 2 = \dfrac{27}{2}$.

Practise with new numbers

Try it in Python

This program checks the laws with numbers, and it finds some fractional powers.

Common mistakes

• Adding the powers when the bases differ. $2^3 \times 3^2$ is not $6^5$.
• Multiplying the powers in the first law. $a^m \times a^n$ adds the powers. $(a^m)^n$ multiplies them.
• Thinking that $a^{-n}$ is negative. $2^{-3} = \dfrac{1}{8}$, which is positive.
• Reading $a^{1/2}$ as half of $a$. It is the square root of $a$.

Remember these

• $a^m a^n = a^{m+n}$, $(a^m)^n = a^{mn}$, $\dfrac{a^m}{a^n} = a^{m-n}$, $a^0 = 1$, $a^{-n} = \dfrac{1}{a^n}$.
• $a^{1/n}$ is the $n$-th root of $a$.
• $a^{m/n} = \left(a^{1/n}\right)^m$. Find the root first, as it keeps the numbers small.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $125^{1/3}$.

2. Find $32^{2/5}$.

3. Simplify $3^{1/2} \times 3^{3/2}$.

4. Simplify $\dfrac{2^{5/3}}{2^{2/3}}$.