Chapter 7, Problem 29PT

To determine

To Simplify:

  $\frac{a^{-\frac{1}{2}}}{6a^{\frac{1}{3}}\cdot a^{-\frac{1}{4}}}$

Answer to Problem 29PT

  $\frac{1}{6a^{\frac{7}{12}}}$

Explanation of Solution

Given Information:

The given expression is :

  $\frac{a^{-\frac{1}{2}}}{6a^{\frac{1}{3}}\cdot a^{-\frac{1}{4}}}$

Concept Used:

• Definition of nth root:

• Words : For any real numbers a and b , and any positive integer n, if $a^n=b$ then a is the nth root of b. That is $a=\sqrt[n]{b}$
• $a^{-n}=\frac{1}{a^n}$
• Product property of radicals : $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}$
• $\frac{a^x}{a^y}=a^{x-y}$
• $a^x\cdot a^y=a^{x+y}$
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• ${\left(\sqrt[n]{a}\right)}^n=a$

Calculation:

  $\begin{array}{l}\frac{a^{-\frac{1}{2}}}{6a^{\frac{1}{3}}\cdot a^{-\frac{1}{4}}}\\ =\frac{a^{-\frac{1}{2}}}{6a^{\frac{1}{3}-\frac{1}{4}}}\mathrm{............}[a^x\cdot a^y=a^{x+y}]\\ =\frac{a^{-\frac{1}{2}}}{6a^{\frac{1}{12}}}\\ =\frac{1}{6}\times a^{-\frac{1}{2}-\frac{1}{12}}\mathrm{.............}[a^{-n}=\frac{1}{a^n}]\\ =\frac{1}{6}\times a^{-\frac{7}{12}}\\ =\frac{1}{6}\times \frac{1}{a^{\frac{7}{12}}}\mathrm{.............}[a^{-n}=\frac{1}{a^n}]\\ =\frac{1}{6a^{\frac{7}{12}}}\end{array}$