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Precalculus Review Materials

Section 1.3 Integer Exponents

For a non-negative integer \(n\text{:}\)
  • The notation \(a^n\) represents \(a\) multiplied by itself \(n\) times. For example, \(5^3 = 5 \times 5 \times 5\text{.}\)
  • If \(a \neq 0\text{,}\) then \(a^0 = 1\) because \(a\) multiplied by itself zero times is a product that has no terms, and a product that has no terms equals \(1\text{.}\) Note: \(0^0\) is not a number; it is an indeterminate form that will be studied in calculus.
  • Multiplying exponents: When multiplying powers of the same base, add the exponents, because:
    \begin{equation*} a^n \times a^m = \underbrace{aa\cdots a}_{n\ \mathrm{times}} \times \underbrace{aa\cdots a}_{m\ \mathrm{times} } = \underbrace{aa\cdots a \times aa\cdots a}_{n+m\ \mathrm{times} } = a^{n+m} \end{equation*}
    When multiplying exponential expressions of different bases but of the same power, multiply by the bases together and raise it to the exponent, because:
    \begin{equation*} a^n\times b^n = \underbrace{aa\cdots a}_{n\ \mathrm{times}} \times \underbrace{bb\cdots b}_{n\ \mathrm{times} } = \underbrace{abab \cdots ab}_{n\ \mathrm{times} } = (ab)^n \end{equation*}
  • Powers of Powers: When raising a power to another power, multiply the exponents, because:
    \begin{equation*} (a^n)^m = \underbrace{a^n a^n\cdots a^n}_{m\ \mathrm{times}}= \underbrace{aaaa\cdots aaa \times aaa\cdots aa}_{nm\ \mathrm{times} } = a^{nm} \end{equation*}
  • Negative exponents are a shorthand for a power of the reciprocal, and only make sense if the base is not zero. That is, for \(n > 0\) and \(a \neq 0\text{,}\)
    \begin{equation*} a^{-n} = \big( a^{-1} \big)^n = \left( \frac{1}{a} \right)^n= \frac{1}{a^n}\text{.} \end{equation*}
    Note that \(a^{-1} = \frac{1}{a}\text{.}\)
  • Dividing exponents: When dividing powers of the same base, subtract the exponents because of the meaning of negative exponents (above):
    \begin{equation*} a^n \div a^m = \frac{a^n}{a^m}= a^n\left(\frac{1}{a^m}\right)= a^n \times a^{-m} = a^{n-m}\text{.} \end{equation*}
  • If you have different bases to different powers, sometimes you can combine by factoring. See below.
  • \(\displaystyle 6^3 \cdot 6^7 = 6^{3+7} = 6^{10}\)
  • \(2^4 \cdot 2^7 = 2^{4+7} = 2^{11}\text{.}\) Notice that \(2^{11}\) can arise other ways; for example \(2^{11} =2^{8+3} = 2^8 \cdot 2^3\)
  • \(\displaystyle (4^{60})^2 = 4^{60 \times 2} = 4^{120}\)
  • \(\displaystyle \frac{1}{343} = \frac{1}{7^3} = 7^{-3}\)
  • \(\displaystyle 9^6 \div 9^2 = 9^{6-2} = 9^{4}\)
  • \(\displaystyle \frac{3^{-4}}{3^{-5}} = 3^{-4- (-5 )} = 3^1 = 3\)
  • \(\displaystyle 4^3 5^3 =(4 \times 5)^3 = 20^{3}\)
  • \(\displaystyle 2^5 \cdot 8^6 = 2^5 \cdot (2^3)^6 = 2^5 \cdot 2^{18} = 2^{23}\)
  • \(\displaystyle 3^2 \cdot 9^{-4} = 3^2 \cdot (3^2)^{-4} = 3^2 \cdot (3^{-8}) = 3^{-6}\)
  • \(\displaystyle (35)^2 \cdot 7^9 = (5\cdot 7)^2 \cdot 7^{9} = 5^2 \cdot 7^{11}\)
  • \(\displaystyle 12^3 \cdot 18^4 = (2\cdot 6)^3 \cdot (3\cdot 6)^4 = 2^3\cdot 6^3\cdot 3^4 \cdot 6^4 = 2^3\cdot 3^4 \cdot 6^7 = 3\cdot 6^3\cdot 6^7= 3\cdot 6^{10}\)

Exercises Practice Problems

Evaluate and put the following expressions into the lowest terms:

1.

\(z^4 \cdot z^{-5}\)
Answer.
\(z^{-1}\)

2.

\((9^3)^4\)
Answer.
\(9^{12}=3^{24}\)

3.

\((2^{-2})^{-4}\)
Answer.
\(2^8\)

4.

\(\frac{q^5}{q^2}\)
Answer.
\(q^3\)

5.

\((-4)^3 \cdot 5^3\)
Answer.
\((-20)^3 = - (20^3) = -20^3\)