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    Home / Algebra Trig Review / Algebra / Exponents
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    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    August 27, 2020

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    Simplify each of the following as much as possible. Show All Solutions Hide All Solutions

    1. \(2{x^4}{y^{ - 3}}{x^{ - 19}} + {y^{\frac{1}{3}}}{y^{ - \,\frac{3}{4}}}\) Show Solution
      All of these problems make use of one or more of the following properties. \begin{array}{rclcrcl}{p^n}{p^m} & = & {p^{n + m}} & \hspace{0.5in} & \displaystyle \frac{{{p^n}}}{{{p^m}}} & = & {p^{n - m}} = \displaystyle \frac{1}{{{p^{m - n}}}}\\ {\left( {{p^n}} \right)^m} & = &{p^{nm}} & & {p^0} & = & 1\,{\mbox{, provided }}p \ne 0\\ {\left( {pq} \right)^n} & = & {p^n}{q^n} & & {\displaystyle \left( {\frac{p}{q}} \right)^n} & = & \displaystyle \frac{{{p^n}}}{{{q^n}}}\\ {p^{ - n}} & = & \displaystyle \frac{1}{{{p^n}}} & & \displaystyle \frac{1}{{{p^{ - n}}}} & = & {p^n}\\ \displaystyle {\left( {\frac{p}{q}} \right)^{ - n}} & = & \displaystyle {\left( {\frac{q}{p}} \right)^n} =\displaystyle \frac{{{q^n}}}{{{p^n}}} & & & & \end{array}

      This particular problem only uses the first property.

      \[2{x^4}{y^{ - 3}}{x^{ - 19}} + {y^{\frac{1}{3}}}{y^{ - \,\frac{3}{4}}} = 2{x^{4 - 19}}{y^{ - 3}} + {y^{\frac{1}{3}\, - \,\frac{3}{4}}} = 2{x^{ - 15}}{y^{ - 3}} + {y^{ - \,\frac{5}{{12}}}}\]

      Remember that the \(y\)’s in the last two terms can’t be combined! You can only combine terms that are products or quotients. Also, while this would be an acceptable and often preferable answer in a calculus class an algebra class would probably want you to get rid of the negative exponents as well. In this case your answer would be.

      \[2{x^4}{y^{ - 3}}{x^{ - 19}} + {y^{\frac{1}{3}}}{y^{ - \,\frac{3}{4}}} = 2{x^{ - 15}}{y^{ - 3}} + {y^{ - \,\frac{5}{{12}}}} = \frac{2}{{{x^{15}}{y^3}}} + \frac{1}{{{y^{\frac{5}{{12}}}}}}\]

      The 2 will stay in the numerator of the first term because it doesn’t have a negative exponent.

    2. \({x^{\frac{3}{5}}}{x^2}{x^{ - \,\frac{1}{2}}}\) Show Solution
      \[{x^{\frac{3}{5}}}{x^2}{x^{ - \,\frac{1}{2}}} = {x^{\frac{3}{5}\, + \,2\, - \,\,\frac{1}{2}}} = {x^{\frac{6}{{10}}\, + \,\frac{{20}}{{10}}\, - \,\,\frac{5}{{10}}}} = {x^{\frac{{21}}{{10}}}}\]

      Not much to this solution other than just adding the exponents.

    3. \(\frac{{x{x^{ - \,\frac{1}{3}}}}}{{2{x^5}}}\) Show Solution
      \[\frac{{x{x^{ - \,\frac{1}{3}}}}}{{2{x^5}}} = \frac{{{x^{\frac{2}{3}}}}}{{2{x^5}}} = \frac{{{x^{\frac{2}{3}}}{x^{ - 5}}}}{2} = \frac{{{x^{ - \,\frac{{13}}{3}}}}}{2} = \frac{1}{2}{x^{ - \,\frac{{13}}{3}}}\]

      Note that you could also have done the following (probably is easier….).

      \[\frac{{x{x^{ - \,\frac{1}{3}}}}}{{2{x^5}}} = \frac{{{x^{ - \,\frac{1}{3}}}}}{{2{x^4}}} = \frac{{{x^{ - \,\frac{1}{3}}}{x^{ - 4}}}}{2} = \frac{{{x^{ - \,\frac{{13}}{3}}}}}{2} = \frac{1}{2}{x^{ - \,\frac{{13}}{3}}}\]

      In the second case I first canceled an \(x\) before doing any simplification.

      In both cases the 2 stays in the denominator. Had I wanted the 2 to come up to the numerator with the \(x\) I would have used \({\left( {2x} \right)^5}\) in the denominator. So, watch parenthesis!

    4. \({\left( {\frac{{2{x^{ - 2}}{x^{\frac{4}{5}}}{y^6}}}{{x + y}}} \right)^{ - 3}}\) Show Solution

      There are a couple of ways to proceed with this problem. I’m going to first simplify the inside of the parenthesis a little. At the same time, I’m going to use the last property above to get rid of the minus sign on the whole thing.

      \[{\left( {\frac{{2{x^{ - 2}}{x^{\frac{4}{5}}}{y^6}}}{{x + y}}} \right)^{ - 3}} = {\left( {\frac{{x + y}}{{2{x^{ - \,\frac{6}{5}}}{y^6}}}} \right)^3}\]

      Now bring the exponent in. Remember that every term (including the 2) needs to get the exponent.

      \[{\left( {\frac{{2{x^{ - 2}}{x^{\frac{4}{5}}}{y^6}}}{{x + y}}} \right)^{ - 3}} = \frac{{{{\left( {x + y} \right)}^3}}}{{{2^3}{{\left( {{x^{ - \,\frac{6}{5}}}} \right)}^3}{{\left( {{y^6}} \right)}^3}}} = \frac{{{{\left( {x + y} \right)}^3}}}{{8{x^{ - \,\frac{{18}}{5}}}{y^{18}}}}\]

      Recall that \({\left( {x + y} \right)^3} \ne {x^3} + {y^3}\) so you can’t go any further with this.

    5. \({\left( {\frac{{{x^{\frac{4}{7}}}{x^{\frac{9}{2}}}{x^{ - \,\frac{{10}}{3}}} - {x^2}{x^{ - 9}}{x^{\frac{1}{2}}}}}{{x + 1}}} \right)^0}\) Show Solution

      Don’t make this one harder than it has to be. Note that the whole thing is raised to the zero power so there is only one property that needs to be used here.

      \[{\left( {\frac{{{x^{\frac{4}{7}}}{x^{\frac{9}{2}}}{x^{ - \,\frac{{10}}{3}}} - {x^2}{x^{ - 9}}{x^{\frac{1}{2}}}}}{{x + 1}}} \right)^0} = 1\]
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