Laws of Exponents

Positive Integer Exponents and the Laws of Exponents

When we multiply a real number b by itself n times, we write the result in exponential form as bⁿ. That is:

$ b^n=\underbrace{b\cdot b\cdot\cdots\cdot b}_{n\text{ times}} $

In the expression bⁿ, b is called the base and n is called the exponent (or power).

It immediately follows from this definition that the basic laws of exponents (also known as exponent rules) apply. For any real numbers b and c, and positive integers m and n:

In the following sections, we explore how to give meaning to b^r when the exponent r is not a positive integer. Mathematical definitions are logically designed so that these five fundamental laws of exponents remain true for all types of numbers.

Zero Exponent Rule (r = 0)

The zero exponent rule states that any non-zero base raised to the power of zero is equal to 1. If $r=0$, we define:

$ b^{0}=1\quad\text{if }b\neq0. $

Why b⁰ = 1 (b ≠ 0) is the only definition consistent with the Product Rule

We want the Product Rule $b^{r}\cdot b^{s}=b^{r+s}$ to hold true when $s=0$. Taking any positive integer $n$ and setting $s=0$, we get: $ b^{n}=b^{n+0}=b^{n}\cdot b^{0}. $ Dividing both sides by $b^{n}$ (which is valid since $b\neq0$): $ b^{0}=1. $ This is the only value of $b^0$ consistent with the fundamental exponent rules. There is no choice involved — the math forces it!

For a detailed explanation of why $0^0$ is often considered undefined, see the FAQ section below.

Fractional Exponents: nth Roots (r = 1/n)

The fractional exponent rule connects exponents to radicals. If r = 1/n (where n is a positive integer), then $b^{1/n}$ is called the $n$th root of $b$. It is the real number $u$ such that $u^{n}=b$. This is also denoted using the radical symbol $\sqrt[n]{b}$:

$ \sqrt[n]{b}=b^{1/n}=u\quad\text{means}\quad u^{n}=b. $

(Note: The square root $\sqrt[2]{b}$ is simply written as $\sqrt{b}$.)

Deriving the Fractional Exponent Rule

We want the Power Rule $(b^{r})^{s}=b^{rs}$ to hold with $r=1/n$ and $s=n$: $ \bigl(b^{1/n}\bigr)^{n}=b^{(1/n)\cdot n}=b^{1}=b. $ So, $b^{1/n}$ must be a number that, when raised to the $n$th power, yields $b$. That is the exact mathematical definition of the $n$th root of $b$.

Whether such a real number u exists, and how many there are, depends on whether the index n is odd or even.

Odd vs. Even Index

• Case 1: n is a positive odd integer. There is exactly one real nth root for each real number b.
• Case 2: $n$ is a positive even integer. Because $u^{n}\geq0$ for all real $u$ and even $n$ (for example, $(-2)^{4}=16>0$), the equation $u^{n}=b$ has no real solution when $b<0$. For $b>0$, the identity $(-u)^{n}=u^{n}$ gives two real $n$th roots: $u$ and $-u$. By convention, the symbol $b^{1/n}$ or $\sqrt[n]{b}$ always denotes the positive (principal) $n$th root.

Sign of $\sqrt[n]{b}$

To summarize the signs of fractional exponents:

• $\sqrt[n]{b}$ is positive if $b>0$.
• $\sqrt[n]{b}$ is negative if $b<0$ and $n$ is odd.
• $\sqrt[n]{b}$ is imaginary (not a real number) if $b<0$ and $n$ is even. (See Section: Principal Square Root of Negative Numbers)

Properties of $n$th Roots

The properties of $n$th roots are not a separate set of rules — they are simply the five laws of exponents applied with $r=1/n$. The table below makes the correspondence explicit.

Property of nth Roots	Corresponding Exponent Law
$(ab)^{1/n}=a^{1/n}b^{1/n}$	Power of a Product Rule with $r=1/n$
$(a/b)^{1/n}=a^{1/n}/b^{1/n}$	Power of a Quotient Rule with $r=1/n$
$(a^{1/n})^{1/m}=a^{1/(mn)}$	Power of a Power Rule with $r=1/n,\,s=1/m$
$(a^{n})^{1/n}=a$ ($n$ odd)	Power Rule with $r=n,\,s=1/n$ (unique root)
$(a^{n})^{1/n}=\lvert a\rvert$ ($n$ even)	Power Rule with $r=n,\,s=1/n$ (positive root)

(We assume throughout that all the roots involved exist as real numbers. Let $a$ and $b$ be real numbers, and $m$ and $n$ be positive integers.)

Proof of $(ab)^{1/n}=a^{1/n}b^{1/n}$ (or $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$)

Let $u=a^{1/n}$ and $v=b^{1/n}$, meaning $u^{n}=a$ and $v^{n}=b$. Using the Power of a Product Rule: $ (uv)^{n}=u^{n}v^{n}=ab. $ Therefore, $uv$ is an $n$th root of $ab$, which proves $a^{1/n}b^{1/n}=(ab)^{1/n}$.

Proof of $\left(\dfrac{a}{b}\right)^{1/n}=\dfrac{a^{1/n}}{b^{1/n}}$ (or $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$)

Let $u=a^{1/n}$ and $v=b^{1/n}$, meaning $u^{n}=a$ and $v^{n}=b$. Using the Power of a Quotient Rule: $ \left(\frac{u}{v}\right)^{n}=\frac{u^{n}}{v^{n}}=\frac{a}{b}. $ Therefore, $u/v$ is an $n$th root of $a/b$, proving $a^{1/n}/b^{1/n}=(a/b)^{1/n}$.

Proof of $\left(a^{1/n}\right)^{1/m}=a^{1/(mn)}$ (or $\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$)

Let $u=\left(a^{1/n}\right)^{1/m}$. By definition, $u^{m}=a^{1/n}$, and $(u^{m})^{n}=a$. Using the Power Rule $(u^{m})^{n}=u^{mn}$: $ u^{mn}=a. $ Therefore, $u$ is an $mn$th root of $a$, proving $\left(a^{1/n}\right)^{1/m}=a^{1/(mn)}$.

Why $\left(a^{n}\right)^{1/n}=a$ when $n$ is odd

We verify that $a$ is an $n$th root of $a^{n}$: $(a)^{n}=a^{n}$. When $n$ is odd, this root is unique, so no absolute value is needed.

Why $\left(a^{n}\right)^{1/n}=|a|$ when $n$ is even

When $n$ is even, both $a$ and $-a$ satisfy $(\pm a)^{n}=a^{n}$. Since the symbol $\sqrt[n]{\cdot}$ strictly denotes the non-negative root, we must use the absolute value $|a|$. For example, $\sqrt{(-3)^{2}}=\sqrt{9}=3=|-3|$, not $-3$.

Important Note: When $n$ is even and both $a<0$ and $b<0$, then $\sqrt[n]{ab}$ and $\sqrt[n]{a/b}$ exist as real numbers (since $ab>0$ and $a/b>0$), but $\sqrt[n]{a}$ and $\sqrt[n]{b}$ individually do not. In real arithmetic, you cannot split the root using the Product and Quotient formulas in this specific case.

Rational Exponent Rule (r = m/n)

The rational exponent rule evaluates fractional powers where the numerator is greater than 1. If r = m/n (where m and n are positive integers and the fraction is simplified to its lowest terms, e.g., reducing 6/4 to 3/2), then $b^{m/n}$ is defined as the $n$th root of the $m$th power of $b$:

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

It can also be computed identically as the mth power of the nth root:

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

(If $n$ is even, we require $b\geq0$ to remain in the real number system).

Why $b^{m/n} = (b^{1/n})^m$ is the right definition

We want the Product Rule $b^r \cdot b^s = b^{r+s}$ to hold for $r = m/n$. Since $b^{1/n}$ is already established, we write $m/n$ as a sum of $m$ copies of $1/n$: $ b^{m/n} = b^{\,\underbrace{1/n\,+\,1/n\,+\,\cdots\,+\,1/n}_{m\ \text{times}}} = \underbrace{b^{1/n}\cdot b^{1/n}\cdots b^{1/n}}_{m\ \text{times}} = \left(b^{1/n}\right)^{m}. $ This proves that $(b^{1/n})^m$ is the only definition consistent with the Product Rule.
Example calculation: $ 8^{2/3}=\left(8^{1/3}\right)^{2}=2^{2}=4 \quad\text{or}\quad 8^{2/3}=\left(8^{2}\right)^{1/3}=64^{1/3}=4. $

Irrational Exponents

What happens when the exponent is an irrational number, like $\pi$ or $\sqrt{2}$?

If $r=\alpha$ is an irrational exponent and $b>0$, then $b^{\alpha}$ is defined by approximating $\alpha$ with a sequence of rational numbers. Since irrational numbers can be approximated to any desired accuracy by terminating decimals (which are fractions), we can use limits.

For example, to compute $3^{\sqrt{2}}$:
Since $\sqrt{2} \approx 1.41421\ldots$, we can evaluate rational approximations:

• $3^{1.4} = 3^{14/10}$
• $3^{1.41} = 3^{141/100}$
• $3^{1.4142} = 3^{14142/10000}$

The exact value of $3^{\sqrt{2}}$ is the mathematical limit of this sequence.

(Note: When $\alpha$ is irrational and $b<0$, the expression $b^{\alpha}$ is not a real number and enters the realm of complex analysis. Alternatively, advanced mathematics often bypasses limits by defining exponents using the natural logarithm: $b^r = e^{r \ln b}$.)

Negative Exponent Rule

The negative exponent rule states that a negative exponent dictates the reciprocal of the base raised to the positive exponent. If $b\neq0$, we define $b^{-r}$ to be $\dfrac{1}{b^{r}}$ whenever $b^{r}$ is defined.

For example:

$ b^{-\frac{1}{2}}=\frac{1}{b^{1/2}}=\frac{1}{\sqrt{b}} $$ b^{-\frac{5}{3}}=\frac{1}{b^{5/3}}=\frac{1}{\sqrt[3]{b^{5}}}=\frac{1}{b\sqrt[3]{b^{2}}} $

Why $b^{-r}=1/b^{r}$ is the only definition consistent with the Product Rule

We want the Product Rule $b^r \cdot b^s = b^{r+s}$ to hold with $s=-r$: $ b^{r}\cdot b^{-r}=b^{r+(-r)}=b^{0}=1. $ Dividing both sides by $b^{r}$ (valid since $b\neq 0$): $ b^{-r}=\frac{1}{b^{r}}. $

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Frequently Asked Questions (FAQs) About Exponent Rules

Why does a negative exponent not make the result negative?

A negative exponent signals a fractional reciprocal, not a negative value. By the Negative Exponent Rule, $b^{-r}=1/b^{r}$. For example, $2^{-3}=1/2^{3}=1/8$, which is a positive number. The minus sign lives in the exponent, dictating division, not the sign of the final value.

When does $\sqrt[n]{ab}=\sqrt[n]{a}\,\sqrt[n]{b}$ fail?

This identity requires both $\sqrt[n]{a}$ and $\sqrt[n]{b}$ to exist as real numbers. When $n$ is even and both $a<0$ and $b<0$, each individual root is imaginary, yet $\sqrt[n]{ab}$ is real (since multiplying two negatives makes a positive $ab>0$). In that scenario, the identity cannot be applied directly using real arithmetic.

Why is $0^{0}$ undefined?

The expression $0^0$ causes a conflict between two fundamental rules of math:

1. The Zero Exponent Rule: $b^{0}=1$ for all non-zero $b$. This suggests $0^{0}$ should equal $1$.
2. 2. The Base Zero Rule: $0^{n}=0$ for all positive exponents $n$. If we calculate $0^r$ using positive numbers closer and closer to zero (e.g., $0^{0.1} = \sqrt[10]{0}=0$, $0^{0.01}=\sqrt[100]{0}=0$), the result is always $0$. This suggests $0^0$ should equal $0$.

Because limits approaching $0^0$ give conflicting answers depending on the direction you approach from, we consider $0^{0}$ undefined. However, it is worth noting that in certain branches of algebra and combinatorics, mathematicians explicitly define $0^0 = 1$ to simplify polynomial formulas (like the binomial theorem). For general arithmetic purposes, it remains undefined.