Exponents and radicals

Exponents are a short way to write repeated multiplication. Radicals, such as square roots, undo exponents. This page reviews the rules for both. It ends with rational exponents, which connect the two ideas.

Integer exponents

If ${n}$ is a positive integer, ${a^n}$ means ${a}$ multiplied by itself ${n}$ times:

$a^n=\underbrace{a\cdot a\cdot\,\cdots\,\cdot a}_{n\text{ factors}}$

The number ${a}$ is the base, and ${n}$ is the exponent (or power). For example, ${2^5=2\cdot 2\cdot 2\cdot 2\cdot 2=32}$. We read ${2^5}$ as "2 to the fifth power."

Zero and negative exponents are defined so that the laws below keep working. For ${a\ne 0}$:

${a^0=1}$   and   ${a^{-n}=\dfrac{1}{a^n}}$

For example, ${5^0=1}$ and ${2^{-3}=\dfrac{1}{2^3}=\dfrac{1}{8}}$. A negative exponent does not make the number negative. It means "one over" the power.

Watch the parentheses with negative bases:

Laws of exponents

These laws hold for any integers ${m}$ and ${n}$, as long as no base in a denominator is ${0}$:

LawExample
${a^m a^n=a^{m+n}}$${x^2x^5=x^{2+5}=x^7}$
${\dfrac{a^m}{a^n}=a^{m-n}}$${\dfrac{x^7}{x^3}=x^{7-3}=x^4}$
${(a^m)^n=a^{mn}}$${(x^3)^4=x^{3\cdot 4}=x^{12}}$
${(ab)^n=a^nb^n}$${(2x)^3=2^3x^3=8x^3}$
$\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$$\left(\dfrac{x}{3}\right)^2=\dfrac{x^2}{9}$

The first two laws apply only when the bases are the same. For example, ${x^2y^3}$ cannot be combined into one power.

Example 1: Simplify. Write each answer with positive exponents only.

(a) ${(3x^2y^4)(2x^5y)}$

(b) ${(2a^3b^{-2})^3}$

(c) ${\dfrac{x^{-3}y^2}{x^2y^{-1}}}$

Solution:

(a) Multiply the numbers. Then add the exponents of each variable. (Here ${y=y^1}$.)

$\begin{align*}&(3x^2y^4)(2x^5y)\\&=(3\cdot 2)\,x^{2+5}\,y^{4+1}\\&=6x^7y^5\end{align*}$

(b) Raise each factor to the third power. Multiply the exponents:

$\begin{align*}&(2a^3b^{-2})^3\\&=2^3a^{3\cdot 3}b^{-2\cdot 3}\\&=8a^9b^{-6}\\&=\dfrac{8a^9}{b^6}\end{align*}$

(c) Subtract the exponents of each variable, top minus bottom:

$\begin{align*}&\dfrac{x^{-3}y^2}{x^2y^{-1}}\\&=x^{-3-2}\,y^{2-(-1)}\\&=x^{-5}y^{3}\\&=\dfrac{y^3}{x^5}\end{align*}$

Scientific notation

Scientists write very large and very small numbers in scientific notation:

${a\times 10^n}$,   where ${1\le a<10}$ and ${n}$ is an integer.

A positive exponent ${n}$ means a large number. A negative exponent means a small number. For example, the speed of light is about ${3\times 10^8}$ m/s, and the mass of a hydrogen atom is about ${1.67\times 10^{-27}}$ kg.

Example 2:

(a) Write ${5{,}300{,}000}$ and ${0.000072}$ in scientific notation.

(b) Find ${(3\times 10^4)(5\times 10^{-7})}$. Write the answer in scientific notation.

Solution:

(a) Move the decimal point until one nonzero digit is left of it. Count the places moved.

  • ${5{,}300{,}000=5.3\times 10^6}$. The point moved ${6}$ places to the left, so the exponent is ${6}$.
  • ${0.000072=7.2\times 10^{-5}}$. The point moved ${5}$ places to the right, so the exponent is ${-5}$.

(b) Multiply the numbers. Then add the exponents of ${10}$:

$\begin{align*}&(3\times 10^4)(5\times 10^{-7})\\&=(3\cdot 5)\times 10^{4+(-7)}\\&=15\times 10^{-3}\end{align*}$

But ${15}$ is not between ${1}$ and ${10}$. Write ${15=1.5\times 10^1}$:

$\begin{align*}&15\times 10^{-3}\\&=1.5\times 10^{1+(-3)}\\&=1.5\times 10^{-2}\end{align*}$

Roots

A square root of a number ${a}$ is a number ${b}$ with ${b^2=a}$. For example, ${5}$ and ${-5}$ are both square roots of ${25}$, because ${5^2=25}$ and ${(-5)^2=25}$.

The symbol ${\sqrt{a}}$ means the principal square root: the one that is not negative. So ${\sqrt{25}=5}$, not ${-5}$.

In the same way, ${\sqrt[n]{a}}$ is the principal ${n}$th root of ${a}$. It is a number ${b}$ with ${b^n=a}$. The number ${n}$ is called the index, and ${a}$ is the radicand. The whole expression is called a radical. For example:

When the index is even, the radicand cannot be negative, because no real number raised to an even power is negative. For example, ${\sqrt{-4}}$ is not a real number. When the index is odd, the radicand can be any real number.

One more fact is easy to get wrong. For any real number ${a}$,

${\sqrt{a^2}=|a|}$.

For example, ${\sqrt{(-3)^2}=\sqrt{9}=3=|-3|}$. The square root is never negative, so we need the absolute value.

Simplifying radicals

These laws hold when the roots are real numbers:

LawExample
${\sqrt[n]{ab}=\sqrt[n]{a}\,\sqrt[n]{b}}$${\sqrt{50}=\sqrt{25}\,\sqrt{2}=5\sqrt{2}}$
$\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$$\sqrt[3]{\dfrac{8}{27}}=\dfrac{\sqrt[3]{8}}{\sqrt[3]{27}}=\dfrac{2}{3}$

To simplify a radical, take out every factor that is a perfect power for the index. Perfect squares are ${4, 9, 16, 25, \ldots}$, and perfect cubes are ${8, 27, 64, \ldots}$.

Example 3: Simplify. Assume ${x>0}$.

(a) ${\sqrt{72}}$

(b) ${\sqrt[3]{16x^4}}$

(c) ${3\sqrt{12}+\sqrt{27}}$

Solution:

(a) The largest perfect square factor of ${72}$ is ${36}$:

${\sqrt{72}=\sqrt{36}\,\sqrt{2}=6\sqrt{2}}$

(b) Look for perfect cubes: ${16=8\cdot 2}$ and ${x^4=x^3\cdot x}$:

$\begin{align*}&\sqrt[3]{16x^4}\\&=\sqrt[3]{8x^3}\,\sqrt[3]{2x}\\&=2x\sqrt[3]{2x}\end{align*}$

(c) Simplify each radical first. Then they have the same radicand, and we can add them like ${6t+3t=9t}$:

$\begin{align*}&3\sqrt{12}+\sqrt{27}\\&=3\cdot 2\sqrt{3}+3\sqrt{3}\\&=6\sqrt{3}+3\sqrt{3}\\&=9\sqrt{3}\end{align*}$

Watch out: radicals with different radicands cannot be added. For example, ${\sqrt{2}+\sqrt{3}}$ is not ${\sqrt{5}}$.

Rationalizing the denominator

An answer like ${\dfrac{1}{\sqrt{3}}}$ is usually rewritten without a radical in the denominator. This is called rationalizing the denominator. We multiply the numerator and the denominator by the same number, so the value does not change.

Example 4: Rationalize the denominator.

(a) ${\dfrac{6}{\sqrt{3}}}$

(b) ${\dfrac{2}{3-\sqrt{2}}}$

Solution:

(a) Multiply the numerator and the denominator by ${\sqrt{3}}$. Use ${\sqrt{3}\cdot\sqrt{3}=3}$:

$\begin{align*}&\dfrac{6}{\sqrt{3}}\\&=\dfrac{6\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\\&=\dfrac{6\sqrt{3}}{3}\\&=2\sqrt{3}\end{align*}$

(b) The denominator has two terms. Multiply by its conjugate, ${3+\sqrt{2}}$: the same two terms with the opposite sign between them. The denominator becomes a difference of squares, ${(a-b)(a+b)=a^2-b^2}$:

$\begin{align*}&\dfrac{2}{3-\sqrt{2}}\\&=\dfrac{2(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}\\&=\dfrac{6+2\sqrt{2}}{9-2}\\&=\dfrac{6+2\sqrt{2}}{7}\end{align*}$

Rational exponents

Exponents can also be fractions. A fraction exponent means a root:

${a^{1/n}=\sqrt[n]{a}}$   and   ${a^{m/n}=\left(\sqrt[n]{a}\right)^m}$

The denominator of the exponent is the index of the root. The numerator is the power. For example, ${9^{1/2}=\sqrt{9}=3}$ and ${8^{2/3}=\left(\sqrt[3]{8}\right)^2=2^2=4}$.

All the laws of exponents work for rational exponents too.

Example 5: Simplify. Assume ${x>0}$.

(a) ${16^{-3/4}}$

(b) ${x^{1/2}\cdot x^{1/3}}$

(c) ${\left(27x^6\right)^{2/3}}$

Solution:

(a) The negative exponent means "one over". Then take the fourth root, and cube it:

$\begin{align*}&16^{-3/4}\\&=\dfrac{1}{16^{3/4}}\\&=\dfrac{1}{\left(\sqrt[4]{16}\right)^3}\\&=\dfrac{1}{2^3}=\dfrac{1}{8}\end{align*}$

(b) Add the exponents. Use the common denominator ${6}$:

$\begin{align*}&x^{1/2}\cdot x^{1/3}\\&=x^{1/2+1/3}\\&=x^{3/6+2/6}\\&=x^{5/6}\end{align*}$

(c) Raise each factor to the ${\tfrac{2}{3}}$ power. Multiply the exponents:

$\begin{align*}&\left(27x^6\right)^{2/3}\\&=27^{2/3}\,x^{6\cdot\frac{2}{3}}\\&=\left(\sqrt[3]{27}\right)^2x^4\\&=9x^4\end{align*}$

Summary

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