Rules of Exponents - Algebraic

For Rules of Exponents applied to numerical examples instead of algebraic expressions, read Rules of Exponents.

Exponents are a shorthand way for us to write repeated multiplication. We can easily find the value of $a^ b$ by multiplying $a$ out many times. For example, with numerous calculations,

$$2 ^2 \times 2 ^ 3 \times 2 ^ 4 = 4 \times 8 \times 16 = 512 = 2 ^ 9 .$$

However, this approach will quickly lead to large numbers, which introduces complications. The rules of exponents offer us a way to shortcut this process, and conclude that

$$2 ^ 2 \times 2^3 \times 2^4 = 2^ { 2+3 + 4 } = 2 ^ 9.$$

The Exponent Rules are:

$$\begin{array} { c | c } \text{ Rule name} & \text{ Rule } \\ \hline \text{Product Rule} & a^m \times a^n = a^{ m + n } \\ & a ^n \times b^n = (a \times b)^ n \\ \hline \text{Quotient Rule} & a^n / a^m = a^ { n - m } \\ & a^n / b^n = (a/b) ^ n \\ \hline \text{Negative Exponent} & a^ {-n} = \frac{1}{a^n} \\ \hline \text{Power Rule} & (a^n)^m = a^ { n \times m } \\ & a ^ { 1/n} = \sqrt[n]{a } \\ & \sqrt[m]{ a^n} = a^ { n/m} \\ \hline \text{Tower Rule} & a ^ { n^ m } = a ^ { \left ( n^ m \right) } \\ \hline \text{Zero Rule} & a^0 = 1 \\ & 0^ a = 0 \text{ for } a > 0 \\ \hline \text{One Rule} & a^1 = a \\ & 1^a = 1 \\ \end{array}$$

These rules are true if $a$ is positive, and $m$, $n$ are positive integers.

Note: For the Power Rule, with $n = 2$ and $m = \dfrac{1}{2}$, the LHS (left-hand side) is $(a^2) ^ { \frac{1}{2} } = | a |$, while the RHS (right-hand side) is $a^ { 2 \times \frac{1}{2} } = a$. These are not equal. There are also special cases to consider when dealing with negative or complex values.

Because most people have difficulty with negative and fractional exponents, we have added additional exercises for these sections.

$$\large a^m \times a^n = a^{ m + n }$$

When the bases are the same, you can add the exponents.

What is $a^3 \times a^4$?

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We have

$$a^3 \times a^4 = (a \times a \times a) \times (a \times a \times a \times a) = a^7.$$

In other words, $$a^3 \times a^4 = a ^{3+4} = a^7. \ _\square$$

What is $x \times x^2 \times x^6 \times x^4$?

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We have

$$x \times x^2 \times x^6 \times x^4 = x^{1+2+6+4} = x^{13}. \ _\square$$

What is $a^2 \times a^3 \times b \times c^2 \times b^4 \times c^7$?

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We have

$$a^2 \times a^3 \times b \times c^2 \times b^4 \times c^7 = a^{2+3}b^{1+4}c^{2+7} = a^5b^5c^9. \ _\square$$

$$\large a ^n \times b^n = (a \times b)^ n$$

When the exponents are the same, you multiply the bases.

What is $(a \times b)^5$?

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We have

$$(a \times b)^5 = a^5 \times b^5 = a^5 b^5. \ _\square$$

What is $a^3 \times b^3$?

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We have

$$a ^ 3 \times b ^ 3 = ( a \times b )^3 = (ab)^3. \ _\square$$

$$\large \frac{a^n}{a^m} = a^ { n - m }$$

What is $a^6 \div a^3$?

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We have

$$a^6 \div a^3 = a ^{6-3} = a^3. \ _\square$$

What is $a^{5} \div a^{7}$?

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We have

$$a^{5} \div a^{7} = a^{5-7} = a^{-2}. \ _\square$$

What is $x^2 \div x^4$?

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We have

$$x^2 \div x^4 = x^{2-4} = x^{-2} = \frac{1}{x^2}. \ _\square$$

$$\large \frac{a^n}{b^n} = \left(\frac{a}{b}\right) ^ n$$

When the exponents are the same, you can divide the bases.

What is $\frac{a^2}{b^2}?$

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We have

$$\frac{a^2}{b^2}= \left( \frac{a}{b} \right)^2. \ _\square$$

A negative exponent occurs when we have a number written $a^b$, where $b$ is less than zero. To interpret this, if we substitute $n = 0$ and $m = 1$ in the quotient rule, we get

$$a^{-1} =a^{0-1}=\frac{a^0}{a^1} = \frac{1}{a}.$$

More generally, we have

$$a ^ { -n } = \frac{ a^ 0 } { a^n } = \frac{ 1 } { a^n} .$$

Reciprocal:

For any non-zero real number $a,$ the number $a^{-1}$ is its reciprocal. $_\square$

What is $a^{-3}$?

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We have

$$a^{-3} = \frac{1}{a^3}. \ _\square$$

The expression $\dfrac{x^5 \cdot y^{-3} }{x^{-2} \cdot y^3}$ can be simplified to $x^a \cdot y^b$. Find $a + b.$

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The given expression can be rewritten as

$$\begin{aligned} \dfrac{x^5 \cdot y^{-3} }{x^{-2} \cdot y^3} &=x^5 \cdot y^{-3} \cdot \left(\frac{1}{x}\right)^{-2} \cdot \left(\frac{1}{y}\right)^3\\ &= x^5 \cdot y^{-3} \cdot x^2 \cdot y^{-3}\\ &= x^5 \cdot x^2 \cdot y^{-3}\cdot y^{-3}\\ &= x^{5+2} \cdot y^{-3 -3}\\ &= x^{7} \cdot y^{-6}. \end{aligned}$$

On comparing, we get $a = 7$ and $b = -6,$ which implies

$$a + b = 7 - 6 =1. \ _\square$$

$$\large (a^m)^n = a^{m \times n }$$

What is $\left(\left(a^2\right)^5\right)^3$?

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We have

$$\left(\left(a^2\right)^5\right)^3 = a^{2 \times 5 \times 3} = a^{30}. \ _\square$$

What is $\left(a^4\right)^2 \times \left(a^2\right)^3$?

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We have

$$\left(a^4\right)^2 \times (a^2)^3 = a^{4 \times 2} \times a^{2 \times 3} = a^{8+6} = a^{14}. \ _\square$$

What is $\left(a^2\right)^5 \times \left(b^4\right)^3$?

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We have

$$\left(a^2\right)^5 \times \left(b^4\right)^3 = a^{2 \times 5} \times b^{4 \times 3} = a^{10}b^{12}. \ _\square$$

What is $\left(a^7 \times b^4\right)^5$?

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We have

$$\left(a^7 \times b^4\right)^5 = a^{7 \times 5} \times b^{4 \times 5} = a^{35} b^{20}. \ _\square$$

What is $\left( \frac{a^3}{b^2} \right)^4$?

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We have

$$\left( \frac{a^3}{b^2} \right)^4 = \frac{a^{3 \times 4}}{b^{2 \times 4}} = \frac{a^{12}}{b^8}. \ _\square$$

$$\large a ^ { n^ m } = a ^ { \left ( n^ m \right) }$$

In a tower of exponents, work from the top down.

This is a very common mistake made, because we forget that the definition is to work form the top down, as opposed to from the bottom up. Taking $a ^ { ( b^ c ) } = \left ( a ^ b \right) ^ c$ is a common mistake that is made.

If $x > 1$, which of the following is larger: $x ^ { 2 ^ 3 }$ or $\left( x ^2 \right) ^ 3?$

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We have $x ^ { 2 ^ 3 } = x ^ 8$, and $\left ( x^2 \right) ^ 3 = x^ { 2 \times 3 } = x ^ 6$. Since $x > 1$, we have $x ^ 8 > x ^ 6$. Hence, $x ^ { 2 ^ 3 }$ is larger. $_\square$

$$\large \begin{array}{c}&a ^ { 1/n} = \sqrt[n]{a }, &a^ { m/n} = \sqrt[n]{ a^m}. \end{array}$$

Raising to a fractional power is similar to taking a root.
The second rule follows by raising the first rule to the $m^{\text{th}}$ power.

What is $x ^ { \frac{ 1 } { 2} }$?

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We have $$x ^ { \frac{ 1 } { 2} } = \sqrt{x} . \ _ \square$$

What is $\left ( x ^ 2 \right) ^ { \frac{ 1 } { 4 } }$?

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We have $$\left( x ^ 2 \right) ^ { \frac{ 1 } { 4 } } = x^ { \frac{ 2 } { 4} } = x ^ { \frac{1}{2} } = \sqrt{x} . \ _\square$$

$$\large 1 ^ a = 1$$

$1$ raised to any power is equal to $1.$

What is $1 ^ x$?

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$1 ^ x$ is always equal to $1$ for any real value of $x$. $_\square$

$$\text{If } a \neq 0, \text{ then } a ^ 0 = 1.\ _\square$$

Any non-zero number raised to the zero power equals $1.$

What is $a^0$, where $a$ is a non-zero real number?

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We have

$$a^0 = 1. \ _\square$$

What is $x^5 \div x^5$, where $x$ is a non-zero real number?

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We have

$$x^5 \div x^5 = x^{5-5} = x^0 = 1. \ _\square$$

For information about $0^0,$ see What is $0^0$.

Simplify

$$\left( \frac{x^5y^{-3}}{x^3y^8} \right)^2.$$

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We have

$$\left( \frac{x^5y^{-3}}{x^3y^8} \right)^2 = \left(x^{5-3}y^{-3-8}\right)^2 = \left(x^2y^{-11}\right)^2 = x^4y^{-22} = \frac{x^4}{y^{22}}. \ _\square$$

Simplify

$$\frac{ \left(27 x^4 y^{-5}\right)^3 }{ \left(81 x^2 y^{-3}\right)^2 }.$$

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We have

$$\begin{aligned} \frac{ \left(27 x^4 y^{-5}\right)^3 }{ \left(81 x^2 y^{-3}\right)^2 } &= \frac{ \left(3^3 x^4 y^{-5}\right)^3 }{ \left(3^4 x^2 y^{-3}\right)^2 } \\ &= \frac{ 3^9 x^{12} y^{-15} }{ 3^8 x^4 y^{-6} } \\ &= 3^{9-8} x^{12-4} y^{-15-(-6)}\\ &= 3^1 x^8 y^{-9} \\ &= \frac{3x^8}{y^9}. \ _\square \end{aligned}$$

If $3^{x-y} = 81$ and $3^{x+y} = 729$, what is $x$?

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Multiply the two expressions together to get the $y$'s to cancel out:

\[ \begin{align} 3^{x-y} \times 3^{x+y} &= 81 \times 729 \\
3^{x-y+x+y} &= 3^4 \times 3^6 \\
3^{2x} &= 3^{10} \\ x &= 5. \ _\square \end{align}\]

If $a^b = a^c,$ where $a$ is a positive number such that $a\ne1,$ then it implies $b = c$.

If $2^8 = x^2$, where $x$ is a positive integer, what is $x?$

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We can rewrite the given equation as

$$2^8 = 2^ {4 \times 2} = \left(2^4\right)^2 = 16^2=x^2,$$

which implies $x = 16.$ $_\square$

If $27^P = 3^7 \times 9^4$, what is $P?$

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First, let's convert all terms to powers of $3:$

$$\begin{aligned} 3 ^ {3P} &= 3^7 \times 3^{4 \times 2}\\ &= 3^7 \times 3 ^ 8\\ &= 3^{7 + 8} \\ &= 3^{15}. \end{aligned}$$

Since the bases are equal, we can say $3P = 15,$ which implies $P = 5.$ $_\square$