## Algebra 1: Common Core (15th Edition)

$\frac{e^2{^0}}{81c^1{^2}}$
You have the expression ($\frac{3c^3}{e^5}$$)^{-4}.Simplify the expression: (\frac{e^5}{3c^3}$$)^{4}$ -flip the numerator and the denominator because $a^{-b}$=$\frac{1}{a^b}$- ($\frac{e^5{^*4}}{3c^3*4}$) -raise the numerator and denominator to power- =$\frac{e^2{^0}}{81c^1{^2}}$