Q. 160 - Intermediate Algebra

Question

In the following exercises,simplify

$$\begin{gathered} \left(a\right) \frac{a^{{\displaystyle \frac{3}{4}}}.a^{-{\displaystyle \frac{1}{4}}}}{a^{-{\displaystyle \frac{10}{4}}}} \\ \left(b\right) {\left(\frac{27 b^{{\displaystyle \frac{2}{3}}}.c^{-{\displaystyle \frac{5}{2}}}}{b^{-{\displaystyle \frac{7}{3}}}.c^{{\displaystyle \frac{1}{2}}}}\right)}^{\frac{1}{3}} \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right) a^3 \\ \left(b\right) \frac{3b}{c} \end{gathered}$$

Part (a) Step 1.Given information

Consider given expression $\frac{a^{{\displaystyle \frac{3}{4}}}.a^{-{\displaystyle \frac{1}{4}}}}{a^{-{\displaystyle \frac{10}{4}}}}$

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

$a^m.a^n=a^{m+n} and \frac{a^m}{a^n}=a^{m-n}$

Part (a) Step 2.Simplify the expression

$$\begin{gathered} \frac{a^{{\displaystyle \frac{3}{4}}}.a^{-{\displaystyle \frac{1}{4}}}}{a^{-{\displaystyle \frac{10}{4}}}}=\frac{a^{{\displaystyle \frac{3}{4}}+\left(-{\displaystyle \frac{1}{4}}\right)}}{a^{-{\displaystyle \frac{10}{4}}}} [Since; a^m.a^n=a^{m+n}] \\ =\frac{a^{{\displaystyle \frac{1}{2}}}}{a^{-{\displaystyle \frac{10}{4}}}} [Add the exponent] \\ =a^{\frac{1}{2}-\left(-\frac{10}{2}\right) (Since; \frac{a^m}{a^n}=a^{m-n}] } \\ =a^{\frac{1+10}{2}} [Simplify the exponents] \\ =a^3 \left[Simplify\right] \\ Therefore,the answer is a^3. \end{gathered}$$

Part (b) Step 1.Given information

Consider the given expression,${\left(\frac{27 b^{{\displaystyle \frac{2}{3}}}.c^{-{\displaystyle \frac{5}{2}}}}{b^{-{\displaystyle \frac{7}{3}}}.c^{{\displaystyle \frac{1}{2}}}}\right)}^{\frac{1}{3}}$

To simplify the expression,by used product and quotient property that is,$a^m.a^n=a^{m+n} and \frac{a^m}{a^n}=a^{m-n}$

Part (b) Step 2.Simplify the expression

$$\begin{gathered} {\left(\frac{27 b^{{\displaystyle \frac{2}{3}}}.c^{-{\displaystyle \frac{5}{2}}}}{b^{-{\displaystyle \frac{7}{3}}}.c^{{\displaystyle \frac{1}{2}}}}\right)}^{\frac{1}{3}}={\left(27\right)}^{\frac{1}{3}}.{\left(b^{\frac{2}{3}-\left(-\frac{7}{3}\right)}\right)}^{\frac{1}{3}}.{\left(c^{-\frac{5}{2}-\frac{1}{2}}\right)}^{\frac{1}{3}} [Since; \frac{a^m}{a^n}=a^{m-n}] \\ ={\left(3^3\right)}^{\frac{1}{3}}.{\left(b\frac{2}{3}-\left(-\frac{7}{3}\right)\right)}^{\frac{1}{3}}.{\left(c^{-\frac{5}{2}-\frac{1}{2}}\right)}^{\frac{1}{3}} [Since; 3^3=27] \\ ={\left(3^3\right)}^{\frac{1}{3}}.{\left(b^3\right)}^{\frac{1}{3}}.{\left(c^{-3}\right)}^{\frac{1}{3}} [Substract the exponent] \\ ={\left(3\right)}^{3\frac{1}{3}}.{\left(b\right)}^{3\frac{1}{3}}.{\left(c\right)}^{-3\frac{1}{3}} [Since; {\left(a^m\right)}^n=a^{mn}] \\ ={\left(3\right)}^{\frac{3}{3}}.{\left(b\right)}^{\frac{3}{3}}.{\left(c\right)}^{\frac{-3}{3}} [Multiplay the exponent] \\ =3.b.c^{-1} [Divide the exponents] \\ =\frac{3b}{c} [Since; a^{-n=}\frac{1}{a^n}] \\ Therefore,the answer is \frac{3b}{c} \end{gathered}$$