## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{16}{25a^{4}b^{26}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(4a^2b^{-4})^3(10a^5b^7)^{-2} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the extended Power 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} (4a^2b^{-4})^3(10a^5b^7)^{-2} \\\\ \left( 4^3a^{2(3)}b^{-4(3)} \right)\left( 10^{-2}a^{5(-2)}b^{7(-2)} \right) \\\\ \left( 4^3a^{6}b^{-12} \right)\left( 10^{-2}a^{-10}b^{-14} \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( 4^3a^{6}b^{-12} \right)\left( 10^{-2}a^{-10}b^{-14} \right) \\\\= \left( 4^3\right)\left( 10^{-2}\right)a^{6+(-10)}b^{-12+(-14)} \\\\= \left( 4^3\right)\left( 10^{-2}\right)a^{-4}b^{-26} .\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} \left( 4^3\right)\left( 10^{-2}\right)a^{-4}b^{-26} \\\\= \dfrac{4^3}{10^{2}a^{4}b^{26}} \\\\= \dfrac{64}{100a^{4}b^{26}} \\\\= \dfrac{\cancel{64}^{16}}{\cancel{100}^{25}a^{4}b^{26}} \\\\= \dfrac{16}{25a^{4}b^{26}} .\end{array}