[tex]d)\,\,\,\dfrac{5^{-5}\cdot 25^{10}}{125^3} = \dfrac{5^{-5}\cdot 5^{2\cdot 10}}{5^{3\cdot 3}} = 5^{-5+20-9} = 5^6\\ \\ \\ e)\,\,\,\dfrac{{(2^3)}^5\cdot {(2^{-6})}^2}{4^2} = \dfrac{2^{3\cdot 5}\cdot 2^{-6\cdot 2}}{2^{2\cdot 2}} = 2^{15-12-4} = 2^{-1} = \dfrac{1}{2}\\ \\\\f)\,\,\, \dfrac{{(3^{-2})}^3\cdot 9^4}{{(3^3)}^3} = \dfrac{3^{-2\cdot 3}\cdot 3^{2\cdot 4}}{3^{3\cdot 3}} = 3^{-6+8-9} = 3^{-7} = \dfrac{1}{3^7}\\ \\ \\g)\,\,\, \dfrac{2^{-3}\cdot 4^3}{8^{-2}} = \dfrac{2^{-3}\cdot 2^{2\cdot 3}}{2^{3\cdot (-2)}} = 2^{-3+6-(-6)} = 2^{9}[/tex]
[tex]h)\,\,\left|\begin{array}{lcl}\dfrac{3^{-12}\cdot 9^5}{27^{-4}\cdot 3^{12}\cdot 9^{-2}} &=& \dfrac{3^{-12}\cdot 3^{2\cdot 5}}{3^{3\cdot (-4)}\cdot 3^{12}\cdot 3^{2\cdot (-2)}} \\\\ &=& 3^{-12+10-(-12)-12-(-4)}\\ \\ &=&3^{2} \\ \\ &=& 9\end{array}\right.[/tex]
