Q. 8.69 - Intermediate Algebra

Question

Simplify:

a. $\frac{m^{{\displaystyle \frac{2}{3}}}·m^{{\displaystyle \frac{-1}{3}}}}{m^{{\displaystyle \frac{-5}{3}}}}$ b. ${\left(\frac{25m^{{\displaystyle \frac{1}{6}}}{n}^{{\displaystyle \frac{11}{6}}}}{m^{{\displaystyle \frac{2}{3}}}{n}^{{\displaystyle \frac{-1}{6}}}}\right)}^{\frac{1}{2}}$

Step-by-Step Solution

Answer

Part (a). The answer is $m^2$

part (b). The answer is  $\frac{5n}{m^{{\displaystyle \frac{1}{4}}}}$.

Part (a) Step 1. Simplifying

Use the Product Property in the numerator, add the exponents.

$$\begin{gathered} \frac{m^{{\displaystyle \frac{2}{3}}}·m^{{\displaystyle \frac{-1}{3}}}}{m^{{\displaystyle \frac{-5}{3}}}} \\ \frac{m^{{\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{3}}}}{m^{{\displaystyle \frac{-5}{3}}}} \\ \frac{m^{{\displaystyle \frac{2-1}{3}}}}{m^{{\displaystyle \frac{-5}{3}}}} \\ \frac{m^{{\displaystyle \frac{1}{3}}}}{m^{{\displaystyle \frac{-5}{3}}}} \end{gathered}$$

Use the Quotient Property, subtract the

exponents,

$$\begin{gathered} \frac{m^{{\displaystyle \frac{1}{3}}}}{m^{{\displaystyle \frac{-5}{3}}}} \\ {m}^{\frac{1}{3}-\left(-\frac{5}{3}\right)} \\ {m}^{\frac{1+5}{3}} \\ {m}^{\frac{6}{3}} \\ {m}^2 \end{gathered}$$

Part (b) Step 1. Simplifying the expression

Follow the order of operations to simplify inside the parentheses first.

Use the Quotient Property, subtract the

exponents.

$$\begin{gathered} {\left(\frac{25m^{{\displaystyle \frac{1}{6}}}{n}^{{\displaystyle \frac{11}{6}}}}{m^{{\displaystyle \frac{2}{3}}}{n}^{{\displaystyle \frac{-1}{6}}}}\right)}^{\frac{1}{2}} \\ {\left(\frac{25m^{{\displaystyle \frac{1}{6}}-{\displaystyle \frac{2}{3}}}}{n^{{\displaystyle \frac{-1}{6}}-{\displaystyle \frac{11}{6}}}}\right)}^{\frac{1}{2}} \\ {\left(\frac{25m^{{\displaystyle \frac{-1}{2}}}}{n^{{\displaystyle -2}}}\right)}^{\frac{1}{2}} \end{gathered}$$

Use the Product to a Power Property,

multiply the exponents

$$\begin{gathered} {\left(\frac{25m^{{\displaystyle \frac{-1}{2}}}}{n^{{\displaystyle -2}}}\right)}^{\frac{1}{2}} \\ \frac{{25}^{{\displaystyle \frac{1}{2}}}·m^{{\displaystyle \frac{-1}{2}}·{\displaystyle \frac{1}{2}}}}{n^{-2·{\displaystyle \frac{1}{2}}}} \\ \frac{5m^{{\displaystyle \frac{-1}{4}}}}{n^{-1}} \\ \frac{5n}{m^{{\displaystyle \frac{1}{4}}}} \end{gathered}$$