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Exponents

Apply the rules of exponents to simplifying expressions.

Applying the Rules of Exponents to Simplify Expressions

Squaring a number is multiplying it by itself, and has that name because it is the area of a square with that side length. Cubing a number is finding the volume of a cube with that length of sides. That’s why we refer to \({5}^{2}\) as five squared, or \({10}^{3}\) as ten cubed. Exponents represent that multiplication.

Let’s remind ourselves of the terminology associated with exponents and what exponents represent. Suppose you want to multiply a number, let’s label that number \(a\), by itself some number of times. Let’s label the number of times \(b\). We denote that as \({a}^{b}\). We say \(a\) raised to the \(b\)th power. When we write or see \({7}^{5}\), we call the 7 the base and we call 5 the exponent. What it represents is 7 multiplied by itself 5 times. This means exponents are used as a shorthand for repeated multiplications, where we write \({7}^{5}=7\times 7\times 7\times 7\times 7\). We would write \({7}^{5}\) and say seven to the fifth power.

The definitions of base and exponent make it possible to understand the exponent rules.

In the previous two sets of rules, we’ve seen exponents applied to products and quotients. Now we look to exponents applied to other exponents. For example, \({({3}^{6})}^{4}={3}^{(6\times 4)}={3}^{24}\). This can be explained by examining what the outer exponent does. We raise \({3}^{6}\) to the fourth power, so we multiply \({3}^{6}\) by itself 4 times, \({({3}^{6})}^{4}={3}^{6}\times {3}^{6}\times {3}^{6}\times {3}^{6}\). Now if we apply the product rule for exponents, this becomes \({3}^{(6+6+6+6)}={3}^{24}\).

Raising an Exponent to an Exponent

Try it.

Expand the following:

  1. \({({6}^{7})}^{3}\)
  2. \({({b}^{12})}^{4}\)

Solution
  1. Using the power rule of exponents, \({({6}^{7})}^{3}={6}^{(7\times 3)}={6}^{21}\).
  2. Using the power rule of exponents, \({({b}^{12})}^{4}={b}^{(12\times 4)}={b}^{48}\).

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

  • Exponents are used to express multiplying a number by itself a number of times. The number being multiplied by itself is the base. The number of times it is multiplied by itself is the exponent, which is often referred to as the power.
  • Understanding that exponents represent repeated multiplication of a base makes it possible to establish some rules for combining exponential expressions, using the product rule, the quotient rule, and the power rule. Additionally, it allows us to formulate distributive rules for exponents.
  • Any non-zero number raised to the 0th power is 1. This makes the definition of the 0th power consistent with the division rule for exponents.
  • For consistency, negative exponents represent the reciprocal of the base raised to the power, so that \({a}^{-n}=\frac{1}{{a}^{n}}\), provided that \(a\ne 0\).

Formulas

  • \[{a}^{n}{a}^{m}={a}^{n+m}\]
  • \[\frac{{a}^{n}}{{a}^{m}}={a}^{(n-m)}\]
  • \({a}^{0}=1\), provided that \(a\ne 0\)
  • \[{(a\times b)}^{n}={a}^{n}\times {b}^{n}\]
  • \[{(\frac{a}{b})}^{n}=\frac{{a}^{n}}{{b}^{n}}\]
  • \[{({a}^{n})}^{m}={a}^{(n\times m)}\]
  • \({a}^{-n}=\frac{1}{{a}^{n}}\), provided that \(a\ne 0\)

Videos

  • Exponential Notation
  • Product and Quotient Rule for Exponents
  • Fraction Raised to a Power
  • Simplifying Expressions with Exponents

Practice (10)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. If possible, use the product rule to simplify the following:

    1. \({21}^{9}\times {21}^{15}\)
    2. \({5}^{9}\times {8}^{4}\)
    Atbildēt uz šo jautājumu
    1. We can apply the product rule to simplify the expression because the bases are the same and we are multiplying. \[{21}^{9}\times {21}^{15}={21}^{(9+15)}={21}^{24}\]
    2. Since the bases are not the same (one is 5, the other 8), this cannot be simplified using the product rule for exponents.

  2. Use the product rule to simplify \({a}^{4}\times {a}^{10}\).

    Atbildēt uz šo jautājumu

    The bases are the same, and we are multiplying, so we apply the multiplication rule to simplify the expression.

    \[{a}^{4}\times {a}^{10}={a}^{(4+10)}={a}^{14}\]

  3. Use the quotient rule to simplify \(\frac{{5}^{19}}{{5}^{11}}\).

    Atbildēt uz šo jautājumu

    We can apply the quotient rule to simplify the expression since the bases are the same and we are dividing.

    \(\frac{{5}^{19}}{{5}^{11}}={5}^{(19-11)}={5}^{8}\)

  4. Use the exponent distributive rule to expand \({(6\times 13)}^{7}\).

    Atbildēt uz šo jautājumu

    Applying the distributive rule to the product, we get \({(6\times 13)}^{7}={6}^{7}\times {13}^{7}\).

  5. Use the exponent distributive rule to expand \({(c\times d)}^{10}\).

    Atbildēt uz šo jautājumu

    Applying the distributive rule to the product, we get \({(c\times d)}^{10}={c}^{10}\times {d}^{10}\).

  6. Use the exponent distributive rule to expand the following:

    1. \({(\frac{4}{9})}^{6}\)
    2. \({(\frac{3}{b})}^{11}\)

    Atbildēt uz šo jautājumu
    1. Applying the distributive rule to the quotient, we get \({(\frac{4}{9})}^{6}=\frac{{4}^{6}}{{9}^{6}}\).
    2. Applying the distributive rule to the quotient, we get \({(\frac{3}{b})}^{11}=\frac{{3}^{11}}{{b}^{11}}\).

  7. Expand the following:

    1. \({({6}^{7})}^{3}\)
    2. \({({b}^{12})}^{4}\)

    Atbildēt uz šo jautājumu
    1. Using the power rule of exponents, \({({6}^{7})}^{3}={6}^{(7\times 3)}={6}^{21}\).
    2. Using the power rule of exponents, \({({b}^{12})}^{4}={b}^{(12\times 4)}={b}^{48}\).

  8. Convert the following to expressions with no negative exponent:

    1. \({3}^{4}\times {5}^{-8}\)
    2. \({a}^{-9}\times {b}^{5}\)
    3. \(\frac{7}{{c}^{-2}}\)

    Atbildēt uz šo jautājumu
    1. Using the negative exponent rule on the \({5}^{-8}\) and multiplying, \({3}^{4}\times {5}^{-8}={3}^{4}\times \frac{1}{{5}^{8}}=\frac{{3}^{4}}{{5}^{8}}\).
    2. Using the negative exponent rule on the \({a}^{-9}\) and multiplying, \({a}^{-9}\times {b}^{5}=\frac{1}{{a}^{9}}\times {b}^{5}=\frac{{b}^{5}}{{a}^{9}}\).
    3. Begin by rewriting the expression as \(\frac{7}{{c}^{-2}}=\frac{7}{1}\times \frac{1}{{c}^{-2}}\). Apply the negative exponent rule to \(\frac{1}{{c}^{-2}}\) in the expression, which becomes \(\frac{7}{1}\times \frac{1}{{c}^{-2}}=7\times {c}^{2}\), which has no negative exponents.

  9. Use negative exponents to rewrite the following expressions with no denominator:

    1. \(\frac{{7}^{3}}{{13}^{9}}\)
    2. \(\frac{{c}^{4}}{{d}^{8}}\)

    Atbildēt uz šo jautājumu
    1. Rewrite the expression \(\frac{{7}^{3}}{{13}^{9}}\) as \(\frac{{7}^{3}}{1}\times \frac{1}{{13}^{9}}\). Then use the definition of negative exponents to rewrite the \(\frac{1}{{13}^{9}}\) as \({13}^{-9}\). Last, multiply, yielding \(\frac{{7}^{3}}{1}\times \frac{1}{{13}^{9}}={7}^{3}\times {13}^{-9}\).
    2. Rewrite the expression \(\frac{{c}^{4}}{{d}^{8}}\) as \(\frac{{c}^{4}}{1}\times \frac{1}{{d}^{8}}\). Then use the definition of negative exponents to rewrite the \(\frac{1}{{d}^{8}}\) as \({d}^{-8}\). Last, multiply, yielding \(\frac{{c}^{4}}{1}\times \frac{1}{{d}^{8}}={c}^{4}\times {d}^{-8}\).

  10. Simplify the following:

    1. \({(\frac{{4}^{2}\times 7}{{9}^{3}})}^{5}\)
    2. \({(\frac{5{a}^{4}}{{b}^{9}})}^{6}\)

    Atbildēt uz šo jautājumu
    1. Step 1: To simplify this, we start by distributing the power 5 across the quotient: \[{(\frac{{4}^{2}\times 7}{{9}^{3}})}^{5}=\frac{{({4}^{2}\times 7)}^{5}}{{({9}^{3})}^{5}}\]

      Step 2: We distribute the power 5 in the numerator across that multiplication: \[{(\frac{{4}^{2}\times 7}{{9}^{3}})}^{5}={\frac{({4}^{2}\times 7)}{{({9}^{3})}^{5}}}^{5}=\frac{{({4}^{2})}^{5}\times {7}^{5}}{{({9}^{3})}^{5}}\]

      Step 3: We apply the power rule where indicated: \[{(\frac{{4}^{2}\times 7}{{9}^{3}})}^{5}={\frac{({4}^{2}\times 7)}{{({9}^{3})}^{5}}}^{5}=\frac{{({4}^{2})}^{5}\times {7}^{5}}{{({9}^{3})}^{5}}=\frac{{4}^{(2\times 5)}\times {7}^{5}}{{9}^{(3\times 5)}}=\frac{{4}^{10}\times {7}^{5}}{{9}^{15}}\]

    2. Step 1: To simplify this, we start by distributing the power 6 across the quotient: \[{(\frac{5{a}^{4}}{{b}^{9}})}^{9}=\frac{{(5\times {a}^{4})}^{6}}{{({b}^{9})}^{6}}\]

      Step 2: We distribute the power 5 in the numerator across that multiplication: \[\frac{{(5\times {a}^{4})}^{6}}{{({b}^{9})}^{6}}=\frac{{(5)}^{6}\times {({a}^{4})}^{6}}{{({b}^{9})}^{6}}\]

      Step 3: We apply the power rule where indicated: \[\frac{{(5)}^{6}\times {({a}^{4})}^{6}}{{({b}^{9})}^{6}}=\frac{{5}^{6}{a}^{24}}{{b}^{54}}\]

Symbols used here

\neq
not equal
The two sides are different.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\prod_{k=1}^{n} a_k
product
Multiply a_k for k = 1 up to n.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.
a \equiv b \pmod n
congruent modulo n
n divides a − b; a and b have the same remainder.
a \mid b,\ \gcd(a,b)
divides, greatest common divisor
b is a multiple of a; the largest number dividing both.
\varphi(n),\ \pi(x)
Euler's totient, prime-counting function
Count of 1..n coprime to n; number of primes up to x.
\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n
integers modulo n
The remainders 0…n−1 with clock arithmetic.
a \bmod n
remainder
What is left after dividing a by n.

How to: Exponents

  1. Apply the rules of exponents to simplifying expressions.
  2. We can apply the product rule to simplify the expression because the bases are the same and we are multiplying.
  3. Since the bases are not the same (one is 5, the other 8), this cannot be simplified using the product rule for exponents.
  4. Applying the distributive rule to the quotient, we get
  5. Applying the distributive rule to the quotient, we get
  6. Using the power rule of exponents,
  7. Using the power rule of exponents,
  8. Using the negative exponent rule on the

Questions people ask

Why are primes so important?

Every integer factors into primes in exactly one way, so primes are the atoms of multiplication. Cryptography relies on that factoring being easy to state and hard to do.

How do I tell whether a big number is prime?

Trial division up to the square root works for small numbers. For large ones, probabilistic tests (Miller–Rabin) give an answer that is wrong with negligible probability, and deterministic tests (AKS) exist but are slower.

Izmēģiniet savu

Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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