Exponent rules

Exponents can make a simple problem look complicated. Once you know the exponent rules, though, many of these problems become straightforward.

Here are the exponent rules you’ll need.

Rule 1: With multiplication, add the exponents

$$3^4\ast 3^5=3^9$$

In this problem, both bases are $3$. When you expand each exponent, you get:

$$3^4\times 3^5=(3\times 3\times 3\times 3)\times (3\times 3\times 3\times 3\times 3)=3^9$$

Both expressions expand into a string of $3$s multiplied together. Altogether, you’re multiplying $3$ by itself $9$ times, so the exponents add: $4+5=9$.

Note that you can only add exponents when the base numbers are the same.

Rule 2: With division, subtract the exponents

$$3^6/3^3=3^3$$

Again, the bases of $3^6$ and $3^3$ are both $3$, so you can subtract the exponents. Conceptually, division cancels matching factors: three of the $3$s in the numerator cancel with the three $3$s in the denominator. The exponents have a difference of $6-3=3$.

$$3^6/3^3=(3\times 3\times 3\times 3\times 3\times 3)/(3\times 3\times 3)=3\times 3\times 3=3^3$$

Try this GRE problem that involves Rule 1 and 2:

$0<x<1$

Quantity;A: $2^4(2^3)x^3/2^5$
Quantity;B: $4^4{x}^5/(4x{)}^3$

Do you know the answer?

Let’s simplify Quantity;A first:

$$\begin{array}{c}{2}^4(2^3)x^3/2^5\\ 2^7{x}^3/2^5\\ 2^2{x}^3\\ 4x^3\end{array}$$

Now simplify Quantity;B:

$$\begin{array}{c}{4}^4{x}^5/(4x{)}^3\\ 4^4{x}^5/((4^3)(x^3))\\ 4^4{x}^2/4^3\\ 4^1{x}^2\\ 4x^2\end{array}$$

Now compare them side by side:

$$\begin{array}{rl}A:& 4x^3\\ B:& 4x^2\end{array}$$

Given $0<x<1$, multiplying by $x$ makes a number smaller. Since $x^3=x^2\cdot x$, you have $x^3<x^2$, so Quantity B is greater.

Rule 3: A negative exponent means an exponent on the other side of a fraction

$$3^{-4}=1/(3^4)$$

To remove a negative exponent, rewrite the expression as a fraction and move the base to the other side of the fraction bar.

A negative exponent does not mean the number itself is negative. It only tells you the factor belongs in the denominator.

Using Rule 2, notice that

• $3^4/3^8=3^{4-8}=3^{-4}$

and canceling four factors of $3$ leaves

• $3^4/3^8=1/3^4$.

Rule 4: Any number to the power of 0 is 1

$$3^0=1$$

Using Rule 2, dividing a power by itself subtracts equal exponents, giving an exponent of $0$.

As a fraction, the numerator and denominator are the same, so the value is $1$:

$$3^3/3^3=3^{3-3}=3^0=1$$

Try this GRE question that involves Rule 3 and 4:

Which of the following represents $y$ in terms;of $z$?

Given: $z^{-4}{z}^4{x}^{-y}=x^y{x}^z$

A. $z$
B. $z/2$
C. $-z/2$
D. $2/z$
E. $-2/z$

Try to solve it, and check your work.

Let’s simplify the expression step by step.

$$\begin{array}{rl}{z}^{-4}{z}^4{x}^{-y}& =x^y{x}^z\\ z^{-4+4}{x}^{-y}& =x^y{x}^z\\ z^0{x}^{-y}& =x^y{x}^z\\ (1)x^{-y}& =x^y{x}^z\\ x^{-y}& =x^y{x}^z\\ 1/x^y& =x^y{x}^z\\ 1& =x^{2y}{x}^z\\ 1/x^z& =x^{2y}\\ x^{-z}& =x^{2y}\\ -z& =2y\\ -z/2& =y\end{array}$$

This shows one way to use negative exponents to rewrite an equation. You can also combine exponents earlier:

$$\begin{array}{rl}{z}^{-4}{z}^4{x}^{-y}& =x^y{x}^z\\ z^{-4+4}{x}^{-y}& =x^y{x}^z\\ z^0{x}^{-y}& =x^y{x}^z\\ x^{-y}& =x^y{x}^z\\ x^{-y}& =x^{y+z}\\ -y& =y+z\\ 0& =2y+z\\ -z& =2y\\ -z/2& =y\end{array}$$

No matter which approach you take, you’ll get the same answer.

Rule 5: When a number with an exponent is raised to another exponent, multiply the exponents

$$(3^3{)}^2=3^6$$

You can rewrite $(3^3{)}^2$ as two copies of $3^3$ multiplied together:

• $(3^3{)}^2=(3\times 3\times 3)\times (3\times 3\times 3)$

That’s six $3$s multiplied in a row, so the result is $3^6$. This matches the exponent rule: $3\times 2=6$.

Be careful with parentheses. $(3^3{)}^2$ is not the same as $3^{(3^2)}$.

$$\begin{gathered} (3^3{)}^2=3^6=729 \\ {3}^{(3^2)}=3^9=19683 \end{gathered}$$

Rule 6: When bases are different with the same exponents, multiply the bases and keep the exponent

$$3^3\times 4^3=12^3$$

This is

• $(3\times 3\times 3)\times (4\times 4\times 4)$

Pair each $3$ with a $4$ to make three factors of $12$:

• $(3\cdot 4)(3\cdot 4)(3\cdot 4)=12\times 12\times 12=12^3$.

Try this GRE question that involves Rule 5 and 6:

Which mixed;number represents this fraction?

$$\frac{3^{2}11^2}{((2^2{)}^2{)}^2}$$

A. $3\frac{25}{256}$
B. $4\frac{65}{256}$
C. $4\frac{155}{256}$
D. $4\frac{235}{256}$
E. $5\frac{65}{256}$

Give it a try and check your work.

Let’s simplify. We’ll write the fraction horizontally so it’s easier to follow:

$$\begin{array}{c}{3}^{2}11^2/((2^2{)}^2{)}^2\\ (3\ast 11{)}^2/((2^2{)}^2{)}^2\\ 33^2/((2^2{)}^2{)}^2\\ 1089/((2^2{)}^2{)}^2\\ 1089/(2^2{)}^{2\ast 2}\\ 1089/(2^2{)}^4\\ 1089/2^{2\ast 4}\\ 1089/2^8\\ 1089/256\\ \approx 4.25390625\end{array}$$

Now interpret the result. Since $4.25390625=4+0.25390625$, only the choices with a whole number of $4$ can be correct, so eliminate choices A and E. The fractional part is less than $0.5$, so eliminate choices C and D (their fractional parts are greater than $0.5$). That leaves choice B.

To double-check, compute the fractional part: $65/256=0.25390625$, which matches.

Turning radicals into exponents

It’s often useful to rewrite radicals as exponents so you can apply exponent rules. A radical can be written as a fractional exponent by using the root as the denominator and the inside exponent as the numerator:

$$\sqrt{25}=\sqrt[2]{25^1}=25^{1/2}$$

Bringing it all together: question walkthrough video

Here’s a video going through one of our practice;questions to demonstrate these ideas in action: