## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{a}{10b^{2}c}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(100ab^3c^2)^{-\frac{1}{2}}(a^6b^{-2})^{\frac{1}{4}} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the Power of the Product Rule of the laws of exponents, which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} (100ab^3c^2)^{-\frac{1}{2}}(a^6b^{-2})^{\frac{1}{4}} \\\\= (10^2ab^3c^2)^{-\frac{1}{2}}(a^6b^{-2})^{\frac{1}{4}} \\\\= \left(10^{2\cdot\left(-\frac{1}{2}\right)}a^{-\frac{1}{2}}b^{3\cdot\left( -\frac{1}{2} \right)}c^{2\cdot\left( -\frac{1}{2} \right)} \right) \left( a^{6\cdot\frac{1}{4}}b^{-2\cdot\frac{1}{4}} \right) \\\\= \left(10^{-\frac{2}{2}}a^{-\frac{1}{2}}b^{-\frac{3}{2} }c^{-1} \right) \left( a^{\frac{6}{4}}b^{-\frac{2}{4}} \right) \\\\= \left(10^{-1}a^{-\frac{1}{2}}b^{-\frac{3}{2} }c^{-1} \right) \left( a^{\frac{3}{2}}b^{-\frac{1}{2}} \right) .\end{array} Using the Product Rule of the laws of exponents, which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left(10^{-1}a^{-\frac{1}{2}}b^{-\frac{3}{2} }c^{-1} \right) \left( a^{\frac{3}{2}}b^{-\frac{1}{2}} \right) \\\\= 10^{-1}a^{-\frac{1}{2}+\frac{3}{2}}b^{-\frac{3}{2}+\left(-\frac{1}{2} \right) }c^{-1} \\\\= 10^{-1}a^{\frac{2}{2}}b^{-\frac{4}{2}}c^{-1} \\\\= 10^{-1}a^{1}b^{-2}c^{-1} .\end{array} Using the Negative Exponent Rule of the laws of exponents, which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 10^{-1}a^{1}b^{-2}c^{-1} \\\\= \dfrac{a^{1}}{10^{1}b^{2}c^{1}} \\\\= \dfrac{a}{10b^{2}c} .\end{array}