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Applying Them Together
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In some of the exercises you will do, there will be multiple steps to simplifying the expression, much like in the Order of Operations. Each of these rules is a tool and all the tools can be used together to simplify expressions.

Rules of Exponents - Applying them Together

Video Source (03:43 mins) | Transcript

Review

  • Product Rule: When the bases are the same, you add the powers.
  • Quotient Rule: When the bases are the same, you subtract the powers.
  • Negative Exponent Rule: Move it to the other part of the fraction (numerator or denominator) and the exponent becomes positive.
  • Power Rule: Multiply the powers.
  • Exponents of 0 and 1: Anything raised to the 1 is itself, anything raised to the 0 is 1.
  • (−1) Raised to an Exponent: If the exponent is even, the answer is positive, if the exponent is odd, the answer is negative.
  • Follow the Order of Operations (PEDMAS).

Additional Resources

Practice Problems

Simplify and Evaluate the following expressions:
  1. \({\text{a}}^{5}\,{\text{b}}^{3}\left ( {\text{a}}\,{\text{b}} \right )^{4}\,{\text{b}} =\) (
    Solution
    x
    Solution:
    \({\text{a}}^{9}\,{\text{b}}^{8}\)
    Details:
    In this example, the first thing we need to do is look at the \(\left ({\text{a}}{\text{b}} \right )^{4}\) part.

    \(\left ({\text{a}}{\text{b}} \right )^{4}\) is the same as \(\text{ab}\) multiplied together 4 times.

    This is an image with two expressions. Top expression is a to the fifth power b to the third power left parentheses a b right parentheses to the fourth power b. The bottom expression is expanded into left parentheses a b right parentheses.left parentheses a b right parentheses left parentheses a b right parentheses left parentheses a b right parentheses.

    This is the same as the variable a multiplied together 4 times and the variable b multiplied together 4 times.

    This is the same image as before but with one more third expression combining the second expression. Top expression is a to the fifth power b to the third power left parentheses a b right parentheses to the fourth power b. The bottom expression is expanded into left parentheses a b right parentheses.left parentheses a b right parentheses left parentheses a b right parentheses left parentheses a b right parentheses. Third expression is left parentheses a a a a b b b b right parentheses.

    Thus, the part \(\left ({\text{a}}{\text{b}} \right )^{4}\) is the same as \({\text{a}}^{4}\,{\text{b}}^{4}\) and we can substitute that back into our original expression.

    This image has the same first expression as before, a to the fifth power b to the third power left parentheses a b right parentheses to the fourth power b. Now with an arrow pointing from the left parentheses a b right parentheses to the fourth power to a to the fourth power b to the fourth power in a second expression of a to the fifth power b to the third power a to the fourth power b to the fourth power b.

    Now we can add the exponents of factors with the same base. Remember that anything that doesn’t have an exponent actually has an invisible exponent of 1. So the variable b on the end has an exponent of 1 and we need to include that in our solution.

    \({\text{a}}^{5}\) and \({\text{a}}^{4}\) become \({\text{a}}^{\left ( 5+4 \right )}={\text{a}}^{9}\)

    \({\text{b}}^{3}\) and \({\text{b}}^{4}\) become \({\text{b}}^{\left ( 3+4+1 \right )}={\text{b}}^{8}\)

    This image has the same first expression as the previous second expression, a to the fifth power b to the third power a to the fourth power b to the fourth power b. Now with three more expression below it in a progressive simplification. Second expression combines like variables into a to the fifth power a to the fourth power b to the third power b to the fourth power b. Third expression is a to the power of left parentheses 5 plus 4 right parentheses b to the power of left parentheses 3 plus 4 plus 1 right parentheses. Last expression is a to the ninth power b to the eighth power.

    Our final solution is \({\text{a}}^{9}\,{\text{b}}^{8}\).
    )
  2. \(\left ({\text{x}}\,{\text{y}} \right )^{3}{\text{x}}\,{\text{y}} =\) (
    Solution
    x
    Solution:
    \({\text{x}}^{4}\,{\text{y}}^{4}\)
    )
  3. \(\dfrac{{\text{x}}^{5}\,{\text{y}}^{3}\,{\text{x}}^{2}}{{\text{x}}^{6}{\text{y}}^{2}}=\) (
    Video Solution
    x
    Solution:
    \({\text{x}}\,{\text{y}}\)
    Details:

    (Video Source | Transcript)
    )
  4. \(\dfrac{{\text{m}}^{3}\,{\text{x}}^{7}}{{\text{m}}^{3}{\text{x}}^{2}}=\) (
    Solution
    x
    Solution:
    \({\text{x}}^{5}\)
    Details:
    In this case, we are using the quotient rule on both variables m and x.

    \(\dfrac{{\color{Blue} {\text{m}}}^{3}\,{\color{DarkOrange} {\text{x}}}^{7}}{{\color{Blue} {\text{m}}}^{3}{\color{DarkOrange} {\text{x}}}^{2}}\)

    First, look at the quotient rule for the m variables.

    \(\dfrac{{\text{m}}^{3}}{{\text{m}}^{3}} = {\text{m}}^{\left ( 3-3 \right )} = {\text{m}}^{0} = 1\)

    This image contains the equation: fraction m to the third power x to the seventh power over m to the third power x to the second power this equals m to the power of left parentheses 3 minus 3 right parentheses times the fraction x to the seventh power over x to the second power. The variable m is tinted blue and the variable x is tinted orange. The dot between the terms in the equation means multiplication.

    Since \({\text{m}}^{\left ( 3-3 \right )} = {\text{m}}^{0} = 1\) and 1 multiplied to anything is itself, we only have the x variables left.

    \(\displaystyle{\color{Blue} {\text{m}}}^{\left ( 3-3 \right )}\cdot\frac{{\text{x}}^{7}}{{\text{x}}^{2}} = {\color{Blue} {\text{m}}}^{0} \cdot\frac{{\text{x}}^{7}}{{\text{x}}^{2}}={\color{Blue} 1}\cdot\frac{{\text{x}}^{7}}{{\text{x}}^{2}}\)

    Now we look at the x variables and their exponents.

    \(\dfrac{{\text{x}}^{7}}{{\text{x}}^{2}} = {\text{x}}^{\left ( 7-2 \right )} = {\text{x}}^{5}\)

    \(\displaystyle{\color{Blue} 1}\cdot\frac{{\color{DarkOrange} {\text{x}}}^{7}}{{\color{DarkOrange} {\text{x}}}^{2}} = {\color{Blue} 1}\cdot {\color{DarkOrange} {\text{x}}}^{\left ( 7-2 \right )} = {\color{Blue} 1}\cdot {\color{DarkOrange} {\text{x}}}^{5} = {\color{DarkOrange} {\text{x}}}^{5}\)

    So we have \(1 \cdot {\text{x}}^{5} = {\text{x}}^{5}\).
    )
  5. \(\left ( {\text{b}}^{4}\,{\text{x}}^{3}\,{\text{y}}\,{\text{b}} \right )^{2}\,{\text{x}} =\) (
    Video Solution
    x
    Solution:
    \({\text{b}}^{10}\,{\text{x}}^{7}\,{\text{y}}^{2}\)
    Details:

    (Video Source | Transcript)
    )
  6. \(\left ( {-}{\text{m}} \right )^{3}\,{\text{b}}^{2}\,{\text{mx}}^{3} =\) (
    Solution
    x
    Solution:
    \(-{\text{m}}^{4}\,{\text{b}}^{2}\,{\text{x}}^{3}\)
    )
  7. \(\left ( -3 \right )^{3}\,{\text{a}}^{2}\,{\text{b}}^{4}\left ( -2 \right )^{2} =\) (
    Solution
    x
    Solution:
    \(-108{\text{a}}^{2}\,{\text{b}}^{4}\)
    )

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(-1) Raised to an Exponent
Introduction to Roots