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- Algebra through Puzzles Supercharge your algebraic intuition and problem solving skills!
Sandeep Bhardwaj, Ashley Toh, Ashish Menon, and
- Andrew Ellinor
- Margaret Zheng
- James Parlier
- Ram Mohith
- Rei N.
- Pi Han Goh
- Parth Lohomi
- Siddhartha Srivastava
- Calvin Lin
- Arron Kau
- Eli Ross
- Jimin Khim
- Zandra Vinegar
- J M
contributed
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.
\[\begin{array} { c | c }\textbf{ Rule name} & \textbf{ Rule } \\ \hline\text{Product Rule} & a^m \times a ^n = a ^{m+n} \\ & a ^n \times b^n = (a \times b)^ n \\\hline\text{Quotient Rule} & a^n / a^m = a^ { n - m } \\ & a^n / b^n = (a/b) ^ n \\\hline\text{Negative Exponent} & a^ {-n} = 1/a^n \\\hline\text{Power Rule} & (a^n)^m = a^ { n \times m } \\[5 pts]\hline\text{Tower Rule}& a ^ { n^ m } = a ^ { \left ( n^ m \right) } \\\hline\text{Fraction Rule} & a ^ { 1/n} = \sqrt[n]{a } \\ & \sqrt[m]{ a^n} = a^ { n/m} \\\hline\text{Zero Rule} & a^0 = 1 \\& 0^ a = 0 \text{ for } a > 0 \\\hline\text{One Rule} & a^1 = a \\& 1^a = 1 \end{array}\]
Contents
- Rule of Exponents: Product
- Rule of Exponents: Quotient
- Negative Exponents
- Power Rule
- Tower of Exponents
- Rule of Exponents: Fractions
- Rule of Exponents: Special Cases
- Problem Solving
- See Also
Rule of Exponents: Product
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 \(2^3 \times 2^4?\)
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 \(3^2 \times 3^6?\)
We have
\[ 3^2 \times 3^6 = 3 ^{2+6} = 3^8. \ _\square \]
\[2000^{2001}\] \[2000^{4000}\] \[4000000^{2000}\] \[4000^{2000}\] \[2000^{4000000}\]
\[\large \color{red}{2000}\left(\color{blue}{2000}^{\color{green}{2000}}\right) = \, \color{purple}? \]
68 70 72 74 75
\[\large \left({\color{blue}3}^{23} - {\color{blue}3}^{22}\right)\left({\color{blue}3}^{24} - {\color{blue}3}^{23}\right) \left({\color{blue}3}^{25} - {\color{blue}3}^{24}\right)\]
This expression can be simplified as \({\color{red}2}^{\color{purple}m} \times {\color{blue}3}^{\color{teal}n}\).
What is \({\color{purple}m} + {\color{teal}n}?\)
\[5^{56}\] \[5^{275}\] \[25^{56}\] \[25^{275}\]
\[\large \color{brown}5^{55} + \color{brown}5^{55} + \color{brown}5^{55} + \color{brown}5^{55} +\color{brown} 5^{55} = \, ? \]
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 \( (2 \times 3)^5?\)
We have
\[ (2 \times 3)^5 = 2^5 \times 3^5 = 2^5 3^5. \ _\square \]
What is \( 2^3 \times 5^3?\)
We have
\[ 2 ^ 3 \times 5 ^ 3 = ( 2 \times 5 )^3 = 10 ^3 = 1000. \ _\square \]
Here is a problem for you to try:
\[ ( 33 \times 33)^3\] \[ (27 \times 33)^3\] \[ (81 \times 33)^3\] \[ ( 333 \times 3)^3\]
\[ \large \left( 3 \times 3^3\right)^3 \times (33)^3 \]
Evaluate the expression above.
Rule of Exponents: Quotient
When the bases of two numbers in division are the 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 \( 2^5 \div 2^3?\)
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 \( \dfrac{ 3^4 }{ 3 ^2} . \)
We have
\[ \frac{ 3^4 }{ 3 ^2} = 3 ^{4-2} = 3^2 =9. \ _\square\]
Practice your mind at the following problems.
\[\dfrac{1}{2}\] \[1\] \[\dfrac{3}{2}\] \[2\]
\[\large \dfrac{ \color{green}2^{\color{blue}{5}}\times \color{green} 2^{\color{blue}{7}}}{\color{green} 2^{{\color{blue}{10}}}\times \color{green} 2^{\color{blue}{3}}}\times \color{purple}{3^0} = \ ? \]
\[0.2\] \[30\] \[36\] \[6^{2005} \] \[6^{2007} \]
\[ \large \frac {\color{red}6^{\color{blue}{2007}} - \color{red}6^{\color{blue}{2006}}}{\color{green}{30}} = \ ? \]
When the exponents of two numbers in division are the 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 \( \left( \dfrac{3}{5} \right)^2?\)
We have
\[ \left( \frac{3}{5} \right)^2 = \frac{3^2}{5^2} = \frac{9}{25}. \ _\square\]
What is \( \left( \dfrac{5^3}{7^2} \right)^4?\)
We have
\[ \left( \frac{5^3}{7^2} \right)^4 = \frac{5^{3 \times 4}}{7^{2 \times 4}} = \frac{5^{12}}{7^8}. \ _\square \]
Let's try the following simple problem:
\[9\] \[9^6\] \[6^9\] \[11^{99}\]
\[\large \dfrac{99^{{6}}}{11^{{6}}} = \, ?\]
Negative Exponents
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 \( 2^{-3}?\)
We have
\[ 2^{-3} = \frac{1}{2^3} = \frac{1}{8}. \ _\square \]
What is \(\dfrac{1}{{10}^{-1}}?\)
We have
\[\begin{align}\dfrac{1}{{10}^{-1}}& = \dfrac{1}{\hspace{2mm} \frac{1}{10}\hspace{2mm} }\\\\& = 10.\ _\square\end{align}\]
What is \( 3^{5} \div 3^{7}?\)
We have
\[ 3^{5} \div 3^{7} = 3^{5-7} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}. \ _\square \]
Simplify \( \dfrac{ 4^{-3} }{ 16^{-1}}. \)
We have
\[\left( \frac{ 4^{-3} }{ 16^{-1}} \right) = \frac{ 4^{-3} }{ \left(4^2\right)^{-1}} = \frac{ 4^{-3} }{ 4^{-2}} = 4^{-3 - (-2)}= 4^{-1}= \frac{1}{4}. \ _\square\]
\[-25\] \[\displaystyle -\frac { 1 }{ 25 }\] \[\displaystyle \frac { 1 }{ 25 }\] \[25\]
\(\displaystyle { \left( -\cfrac { 1 }{ 125 } \right) }^{ -\frac{ 2 }{ 3 } }\) can be written as \(\text{__________}.\)
\[ \large \left [ \left ( \frac 1 3 \right )^{- \frac 4 3} \right]^{\frac 9 2} \times 9^{-3} = \ ? \]
\[\cfrac { 1 }{ 3 } \] \[3\] \[81\] \[{ 81 }^{ 4 }\] None of the above
\[ \Large \left ( \color{purple}{81} \right )^{ - \left (\color{green}2^{- \color{blue}2} \right ) } = \ ? \]
Power Rule
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 \( \big(2^3\big)^4?\)
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 \(2^5 \times 4^3?\)
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 \( \big(2^3\big)^2 \times \big(2^2\big)^5?\)
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:
\[p\] \[p^{1.2}\] \[p^{2.4}\] \[p^{24}\]
\[\large \left[ \left(p^{0.2} \right)^4\right]^3\]
Simplify the expression above for non-negative \(p\).
Tower of Exponents
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, \( a ^ {\large( b^ c ) } = \left ( a ^ b \right) ^ c \) 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 \( 4 ^ { 3 ^ 2 }?\)
We have
\[ 4 ^ { 3 ^ 2 } = 4 ^ 9 = 262144 . \ _\square \]
Note: It is not equal to \( \big( 4 ^ 3 \big) ^ 2 = 64 ^ 2 = 4096 \).
True False
True or False?
For positive reals \(a, b, c,\)
\[ \large \left( a^ {\color{red}b}\right) ^ {\color{blue}c} = \left( a ^ {\color{blue}c} \right)^ {\color{red}b}. \]
64 512 512 or 64 512 and 64 None of the above
\[\Large{2^{3^2}}= \, ?\]
\[\frac{111}{2}\] \[7^{7}-7\] \[7^{42}\] \[7^{70}\] \[7^{7^{7}-14}\]
Which of the following is equal to \[ \large \dfrac{7^{7^{7}}}{(7 \times 7)^{7}} ? \]
\[3^{5^3}\] \[6^{2^6}\] They are equal
Which is greater,
\[\Huge \color{blue}{3}^{\color{red}{5}^{\color{blue}{3}}} \qquad \text{or} \qquad \color{green}{6}^{\color{purple}{2}^{\color{green}{6}}} \color{black}\ ?\]
Rule of Exponents: Fractions
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 \( 81 ^ \frac12?\)
We have
\[ 81 ^ \frac12 = \sqrt{81} = 9 . \ _\square\]
What is \( 343 ^ \frac {2}{3}?\)
We have
\[ 343 ^ { \frac{2}{3} } = \sqrt[3] { 343 ^ 2 } = \sqrt[3] { \big( 7 ^ 3\big) ^ 2 } = \sqrt[3]{ 7 ^ 6 } = 7 ^ { \frac{ 6}{3} } = 7 ^ 2 . \ _\square \]
\[{\huge{16^{\color{RED}{ 0.25}} + 25^{\color{RED} {0.5}}}} \]
Find the value of the expression above.
\[ \Large{ 3 }^{ 4{ x } }\] \[ \Large{ 3 }^{ 8{ x }^{ 2 } }\] \[ \Large{ 9 }^{ 4{ x } }\] \[\Large { 9 }^{ 8{ x }^{ 2 } }\]
\[ \Huge \sqrt{\color{red}9^{\color{green}{16} \color{blue}{x^2}}} \]
Which of the following is equal to the above expression?
Rule of Exponents: Special Cases
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 \( 1 ^ {2345}?\)
We have
\[ 1 ^ { 2345} = 1. \ _\square \]
Note: We do not need to multiply it out \(2345\) times, to know that the product is still \(1.\)
\[\Huge 1^{2^{3^{4^{5^{6^{7^{8}}}}}}} = \ ? \]
What is \( 3^0?\)
We have
\[ 3^0 = 1. \ _ \square \]
0 1 16 None of the above
\[ \LARGE 2^{2^{2^{2^0}}} = \ ? \]
Problem Solving
What is \( 3^{6} \div 3^{4} - 1?\)
Using our knowledge of the order of operations, we divide first, and then subtract:
\[ 3^{6} \div 3^{4} - 1 = \frac{3^6}{3^4} - 1 = 3^{6-4} - 1 = 3^2 - 1 = 9 -1 = 8. \ _\square \]
\[\Large {{\left(x\sqrt[4]{x}\right)}^x = x^{x\sqrt[4]{x}}}\]
Given that \(x\, (\ne 1,0,-1)\) satisfies the equation above and \(x \) can be expressed as \( \frac ab,\) where \(a\) and \(b\) are coprime positive integers, find \(a+b.\)
This is one part of the set Fun with exponents.
\[ 2^3 + 3^4 \] \[ 2^4 + 3^5 \] \[ 2^6 + 3^{12} \] \[ 4^5 \]
\[ \Large 2^{\color{red}3} + 2^{\color{red}3} + 3^{\color{blue}4} + 3^{\color{blue}4} + 3^{\color{blue}4} = {\ \color{teal}} ? \]
\[0\] \[1\] \[\frac { 3 }{ 2 } \] \[3\]
Solve for \(x:\)\[\Large { 3 }^{ 1-\color{red}{x} }=\frac { 1 }{ \sqrt { 3 } }. \]
\[{14}^{14}\] \[{14}^{{14}^{8}}\] \[{14}^{{14}^{13}}\] Cannot be solved
What is the \(14^\text{th}\) root of \({14}^{{14}^{14}}?\)
\[\begin{eqnarray}X&=&a^1\cdot b^3 \cdot c^5\cdot \cdot \cdot \cdot z^{51} \\ Y&=&z^1 \cdot y^3 \cdot x^5 \cdot \cdot \cdot \cdot a^{51} \end{eqnarray}\]
If \(a \cdot b \cdot c \cdot d \cdot \cdot \cdot \cdot z=(-24389)^{\frac{1}{156}}\), then find the value of \(XY\).
\[ \huge 27^{- \frac {x}{3} } + 81^{ \frac {1-x}{4} } \]
If the expression above can be stated in the form of \( \dfrac {a}{b^{^{\large{x}}}} \) for positive integers \(a\) and \(b\), what is the value of \(a+b?\)
\[ \begin{eqnarray}\large \color{purple} {3^x} & = & \color{blue}{2^{2015}} \\\\\large \color{blue}{2^{2015y}} & = & 81 \\\\\large \color{green}{xy} & = & \ \color{teal} ?\end{eqnarray} \]
\[\Large \big({\sqrt[3]{x}}\big)^{x^3} = {x^{3\sqrt[3]{x}}}\]
Find the real value of \(x\) which satisfies the equation above.
If the answer is of the form \(\sqrt[8]{a}\), submit the value of \(a\).
Note: Here \(x \neq 1,0,-1\).
True False Ambiguous Syntax error
\[\huge \color{green}{2^{3^{ \left( 2^{3^{2}} \right)^{0^{2^{3}}}}}}=\color{red}{2^{3^{1^{2^{3^{2^{0^{2^{3}}}}}}}}}\]
Is the equation above true or false?
\[\large {\left(\sqrt[4]{x}\right)}^{4x^4} = {\left(x^4\right)}^{4\sqrt[4]{x}}\]
Find the real value of \(x\) that satisfies the equation above.
If the answer is of the form \(a × \sqrt[15]{b},\) then submit the value of \(a + b.\)
Note: Here, \(x \neq 0,1\).
This is one part of the set Fun with exponents.
\[\large{ \begin{cases}
ab = a^b \\\\\frac{a}{b} = a ^ {3b} \\ \end{cases}} \]
\( a\) and \(b\) are real numbers such that \( a > 1 \) and \( b \neq 0 \), satisfying the above system.
Find \( b ^ { -a}. \)
See Also
To learn more about exponents, check out these related pages:
- Rules of Exponents - Algebraic
- Exponential Functions
- Rational Exponents
- Order of Operations - Exponents
Cite as: Rules of Exponents. Brilliant.org. Retrieved from https://brilliant.org/wiki/simplify-exponents/
