Intermediate Algebra (12th Edition)

$\dfrac{1,250x^{6}z^{7}}{9}$
Use the laws of exponents to simplify the given expression, $\dfrac{(5z^2x^3)^2(2zx^2)^{-1}}{(-10zx^{-3})^{-2}(3z^{-1}x^{-4})^2} .$ Using the law of exponents which states that $(a^xb^y)^z=a^{xz}b^{yz},$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{(5^2z^{2(2)}x^{3(2)})(2^{-1}z^{-1}x^{2(-1)})}{(-10)^{-2}z^{-2}x^{-3(-2)}(3)^2z^{-1(2)}x^{-4(2)}} \\\\= \dfrac{(5^2z^{4}x^{6})(2^{-1}z^{-1}x^{-2})}{(-10)^{-2}z^{-2}x^{6}(3)^2z^{-2}x^{-8}} .\end{array} Using the law of exponents which states that $a^x\cdot a^y=a^{x+y},$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{(5^2)(2^{-1})z^{4+(-1)}x^{6+(-2)}}{(-10)^{-2}(3)^2z^{-2+(-2)}x^{6+(-8)}} \\\\= \dfrac{(5^2)(2^{-1})z^{4-1}x^{6-2}}{(-10)^{-2}(3)^2z^{-2-2}x^{6-8}} \\\\= \dfrac{(5^2)(2^{-1})z^{3}x^{4}}{(-10)^{-2}(3)^2z^{-4}x^{-2}} .\end{array} Using the law of exponents which states that $\dfrac{a^x}{a^y}=a^{x-y},$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{(5^2)(2^{-1})x^{4-(-2)z^{3-(-4)}}}{(-10)^{-2}(3)^2} \\\\= \dfrac{(5^2)(2^{-1})x^{4+2}z^{3+4}}{(-10)^{-2}(3)^2} \\\\= \dfrac{(5^2)(2^{-1})x^{6}z^{7}}{(-10)^{-2}(3)^2} .\end{array} Using the law of exponents which states that $a^{-x}=\dfrac{1}{a^x}$ or $\dfrac{1}{a^{-x}}=a^x$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{(5^2)(-10)^{2}x^{6}z^{7}}{(3)^2(2^{1})} \\\\= \dfrac{(25)(100)x^{6}z^{7}}{(9)(2)} \\\\= \dfrac{(25)(\cancel{100}^{50})x^{6}z^{7}}{(9)(\cancel{2})} \\\\= \dfrac{1,250x^{6}z^{7}}{9} .\end{array}