Q. 162 - Intermediate Algebra

Question

In the following exercises,simplify

$$\begin{gathered} \left(a\right) \frac{m^{{\displaystyle \frac{7}{4}}}.m^{-{\displaystyle \frac{5}{4}}}}{m^{-{\displaystyle \frac{2}{4}}}} \\ \left(b\right) {\left(\frac{16 m^{{\displaystyle \frac{1}{5}}}.n^{{\displaystyle \frac{3}{2}}}}{81 m^{{\displaystyle \frac{9}{5}}}.n^{-{\displaystyle \frac{1}{2}}}}\right)}^{\frac{1}{4}} \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right) m \\ \left(b\right) \frac{2}{3}{m}^{-\frac{2}{5}}.n^{\frac{1}{2}} \end{gathered}$$

Part (a) Step 1.Given information

Consider the given expression,$\frac{m^{{\displaystyle \frac{7}{4}}}.m^{-{\displaystyle \frac{5}{4}}}}{m^{-{\displaystyle \frac{2}{4}}}}$

To simplify the expression,by used product and quotient property that is,

$a^m.a^n=a^{m+n} and \left(\frac{a^m}{a^n}\right)=a^{m-n}$

Part (a) Step 2.Simplify the expression

$$\begin{gathered} \frac{m^{{\displaystyle \frac{7}{4}}}.m^{-{\displaystyle \frac{5}{4}}}}{m^{-{\displaystyle \frac{2}{4}}}}=\frac{m^{{\displaystyle \frac{7}{4}}+\left(-{\displaystyle \frac{5}{4}}\right)}}{m^{-{\displaystyle \frac{2}{4}}}} [Since ; a^m.a^n=a^{m+n}] \\ =\frac{m^{{\displaystyle \frac{1}{2}}}}{m^{-{\displaystyle \frac{2}{4}}}} [Add the exponents] \\ =m^{\frac{1}{2}-\left(-\frac{2}{4}\right)} [Since ; \frac{a^m}{a^n}=a^{m-n}] \\ =m^1 [Simplify the exponents] \\ =m \left[simplify\right] \\ Therefore,the answer is m. \end{gathered}$$

Part (b) Step 1 Given information

Consider the given expression,${\left(\frac{16 m^{{\displaystyle \frac{1}{5}}}.n^{{\displaystyle \frac{3}{2}}}}{81 m^{{\displaystyle \frac{9}{5}}}.n^{-{\displaystyle \frac{1}{2}}}}\right)}^{\frac{1}{4}}$

To simplify the expression,by used product and quotient property that is

$a^m.a^n=a^{m+n} and \left(\frac{a^m}{a^n}\right)=a^{m-n}$

Part (b) Step 2.Simplify the expression