Paul's Online Notes
Paul's Online Notes
Home / Algebra Trig Review / Algebra / Exponents
Show General Notice Show Mobile Notice Show All Notes Hide All Notes
General Notice

Small update to the site today that put the "Next Section" and "Previous Section" buttons in a slightly more obvious place. If they appear below (and slightly overlapping) the "Notes", "Practice Problems" and "Assigment Problems" buttons please clear your browsers cache. Some browsers (Chrome I'm looking at you.....) do not always look to the server to see if newer versions of some of the "background" files are available. Clearing your browsers cache will force them to get the newer versions.

January 27, 2020

Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.


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.

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