Rules of Exponents

For rules of exponents applied to algebraic functions instead of numerical examples, read Rules of Exponents - Algebraic.

To evaluate expressions with exponents, refer to the rules of exponents in the table below. Remember that these rules are true if \(a\) is positive, and \(m\) and \(n\) are real numbers.

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

When the bases of two numbers in multiplication are the same, their exponents are added and the base remains the same. If \(a\) is a positive real number and \(m,n\) are any real numbers, then

\[\large a^m \times a^n = a^{ m + n } .\]

What is \(2^3 \times 2^4\)?

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We have

\[ 2^3 \times 2^4 = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 2^7.\]

In other words,

\[ 2^3 \times 2^4 = 2 ^{3+4} = 2^7. \ _\square \]

What is \(3^2 \times 3^6\)?

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We have

\[ 3^2 \times 3^6 = 3 ^{2+6} = 3^8. \ _\square \]

When the exponents of two numbers in multiplication are the same, then bases are multiplied and the exponent remains the same. If \(a, b\) are positive real numbers and \(n\) is any real number, then we have

\[\large a ^n \times b^n = (a \times b)^ n. \]

Here are some examples based on the above rule.

What is \( (2 \times 3)^5 \)?

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We have

\[ (2 \times 3)^5 = 2^5 \times 3^5 = 2^5 3^5. \ _\square \]

What is \( 2^3 \times 5^3 \)?

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We have

\[ 2 ^ 3 \times 5 ^ 3 = ( 2 \times 5 )^3 = 10 ^3 = 1000. \ _\square \]

Here is a problem for you to try.

When the bases of two numbers in division are same, then exponents are subtracted and the base remains the same. If is \(a\) positive real number and \(m,n\) are any real numbers, then we have

\[\large \dfrac{a^n}{a^m} = a^ { n - m }. \]

Go through the following examples to understand this rule.

What is \( 2^5 \div 2^3 \)?

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We have

\[ 2^5 \div 2^3 = \frac{2 \times 2 \times 2 \times 2 \times 2}{2 \times 2 \times 2} = 2 \times 2 = 2^2. \]

In other words,

\[ 2^5 \div 2^3 = 2^{5-3} = 2^2. \ _\square\]

Simplify \( \dfrac{ 3^4 }{ 3 ^2} . \)

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We have

\[ \frac{ 3^4 }{ 3 ^2} = 3 ^{4-2} = 3^2 =9. \ _\square\]

Practice your mind at the following problems.

When the exponents of two numbers in division are same, then the bases are divided and the exponent remains the same. If \(a, b\) are positive real numbers and \(n\) is any real number, then we have

\[\large \frac{a^n}{b^n} = \left(\frac ab\right) ^ n.\]

Here are the examples to demonstrate the above.

What is \( \left( \dfrac{3}{5} \right)^2 \)?

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We have

\[ \left( \frac{3}{5} \right)^2 = \frac{3^2}{5^2} = \frac{9}{25}. \ _\square\]

What is \( \left( \dfrac{5^3}{7^2} \right)^4 \)?

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We have

\[ \left( \frac{5^3}{7^2} \right)^4 = \frac{5^{3 \times 4}}{7^{2 \times 4}} = \frac{5^{12}}{7^8}. \ _\square \]

Lets try the following simple problem.

For any nonzero real number \( a,\) a negative exponent is handled like so: \( a^{-n} = \dfrac{1}{a^n}. \) Here, the fraction \( \dfrac{1}{a^n} \) is called the reciprocal of \( a^n: \)

\[\large a ^ { -n } = \frac{ 1} { a^n }. \]

In written words, we say "\( a \) to the negative \( n\) equals the reciprocal of \(a \) to the \( n.\)"

What is \( 2^{-3} \)?

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We have

\[ 2^{-3} = \frac{1}{2^3} = \frac{1}{8}. \ _\square \]

What is \(\dfrac{1}{{10}^{-1}}\)?

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We have

\[\begin{align} \dfrac{1}{{10}^{-1}}& = \dfrac{1}{\frac{1}{10}}\\\\ & = 10.\ _\square \end{align}\]

What is \( 3^{5} \div 3^{7} \)?

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We have

\[ 3^{5} \div 3^{7} = 3^{5-7} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}. \ _\square \]

Simplify \( \dfrac{ 4^{-3} }{ 16^{-1}}. \)

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We have \[ \left( \frac{ 4^{-3} }{ 16^{-1}} \right) = \frac{ 4^{-3} }{ \left(4^2\right)^{-1}} = \frac{ 4^{-3} }{ 4^{-2}} = 4^{-3 - (-2)} = 4^{-1} = \frac{1}{4}. \ _\square \]

Power rule of exponents is stated as

\[\large (a^n)^m = a^ { n \times m }. \]

Caution

For the power rule, with \( n = 2 \) and \( m = \frac{1}{2} \), the LHS is \( \big(a^2\big) ^ { \frac{1}{2} } = | a | \), while the RHS 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.

Now these are the examples based on the above rule.

What is \( \big(2^3\big)^4 \)?

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We have

\[ \big(2^3\big)^4 = 2^3 \times 2^3 \times 2^3 \times 2^3 = 2^{3+3+3+3} = 2^{12} .\]

In other words,

\[ \big(2^3\big)^4 = 2^{3 \times 4} = 2^{12}. \ _\square \]

What is \(2^5 \times 4^3\)?

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We have

\[ \begin{align} 2^5 \times 4^3 &= 2^5 \times \big(2^2\big)^3 \\ &= 2^5 \times 2^6 \\ &= 2^{11}. \ _\square \end{align}\]

What is \( \big(2^3\big)^2 \times \big(2^2\big)^5 \)?

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We have

\[ \big(2^3\big)^2 \times \big(2^2\big)^5 = 2^{3 \times 2} \times 2^{2 \times 5} = 2^6 \times 2^{10} = 2^{16} . \ _\square \]

Here is a problem to try.

In a tower of exponents, we work from the top down. So a tower of exponents is evaluated by

\[\large a ^ { n^ m } = a ^ { ( n^ m ) }, \]

an ordering that follows somewhat naturally from the order of operations on exponents.

Towers of exponents problems lend themselves to a common misconception due to an order of operations error. In general, \( a ^ { \left ( b^ c \right ) } = \left ( a ^ b \right) ^ c \) is false. Rare cases wherein the statement is true occur when \( c = 1 \) or \( a = 1 \). One non-trivial example is when \( a = b =c = 2 \). In this case, we have

\[\begin{align} \text{LHS}&: &2 ^ { ( 2 ^ 2 ) } &= 2 ^ 4 = 16 \\ \text{RHS}&: &\big( 2 ^ 2 \big) ^ 2 &= 4 ^ 2 = 16. \end{align} \]

Go through the following examples to understand this rule.

What is \( 4 ^ { 3 ^ 2 } \)?

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We have

\[ 4 ^ { 3 ^ 2 } = 4 ^ 9 = 262144 . \ _\square \]

Note: It is not equal to \( \big( 4 ^ 3 \big) ^ 2 = 64 ^ 2 = 4096 \).

Rule of exponents for fractions works in two steps as

\[\large \begin{array} &a ^ {\frac 1n} = \sqrt[n]{a }, &a^ {\frac mn} = \sqrt[n]{ a^m} \end{array}.\]

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

What is \( 81 ^ \frac12 \)?

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We have

\[ 81 ^ \frac12 = \sqrt{81} = 9 . \ _\square\]

What is \( 343 ^ \frac {2}{3} \)?

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We have

\[ 343 ^ { \frac{2}{3} } = \sqrt[3] { 343 ^ 2 } = \sqrt[3] { \big( 7 ^ 3\big) ^ 2 } = \sqrt[3]{ 7 ^ 6 } = 7 ^ { \frac{ 6}{3} } = 7 ^ 2 . \ _\square \]

It's interesting to note that 1 raised to any power is equal to 1. That is, for any real number \(a\) it is always true that

\[ \large 1 ^ a = 1. \]

Furthermore, any non-zero number raised to the zero power is 1:

\[\large a^0 = 1. \]

NOTE: For \(0^0,\) see what is \(0^0\).

What is \( 1 ^ {2345} \)?

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We have

\[ 1 ^ { 2345} = 1. \ _\square \]

Note: We do not need to multiply it out \(2345\) times, to know that the product is still \(1.\)

What is \( 3^0 \)?

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We have

\[ 3^0 = 1. \ _ \square \]

What is \( 3^{6} \div 3^{4} - 1\)?

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Using our knowledge of the order of operations, we divide first, and then subtract:

\[ 3^{6} \div 3^{4} - 1 = \frac{3^6}{3^4} - 1 = 3^{6-4} - 1 = 3^2 - 1 = 9 -1 = 8. \ _\square \]

To learn more about exponents, check out these related pages:

• Rules of Exponents - Algebraic
• Exponential Functions
• Rational Exponents
• Order of Operations - Exponents