How do you evaluate and simplify #9/9^(-4/5)#?
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"How do you name compounds containing polyatomic ions?"
See a solution process below:
First, use this rule of exponents to eliminate the negative exponent:
#1/x^color(red)(a) = x^color(red)(-a)#
#9/9^color(red)(-4/5) = 9 * 9^color(red)(- -4/5) = 9 * 9^(4/5)#
Next, use these rules to combine the 9's terms:
#a = a^color(red)(1)# and #x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))#
#9 * 9^(4/5) => 9^color(red)(1) * 9^color(blue)(4/5) => 9^(color(red)(1)+color(blue)(4/5)) => 9^(color(red)(5/5)+color(blue)(4/5)) =>#
#9^(9/5)#
#:.color(blue)(=52.196# to the nearest 3 decimal places
#9/(1/9^(-4/5))#
#:.1/a^-2=a^2#
#:.=9/(1/9^(4/5))#
#:.=9/1 xx 9^(4/5)/1#
#:.=9^(5/5) xx 9^(4/5)#
#:.=9^(9/5)#
#:.=root5(9^9)#
#:.=root5(9*9*9*9*9*9*9*9*9)#
#:.root5(9)*root5(9)*root5(9)*root5(9)*root5(9)=9#
#:.=9root5(9*9*9*9)#
#:.=9root5(3*3*3*3*3*3*3*3)#
#:.root5(3)*root5(3)*root5(3)*root5(3)*root5(3)=3#
#:.=3*9root5(27)#
#:.=27root5(27)#
#:.=52.19591521#
#:.color(blue)(=52.196# to the nearest 3 decimal places
