How do you evaluate #9^(3/2)#?

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Answer 1 · 2018-04-13T08:07:26

See below

#9^(3/2)=sqrt( 9·9·9)=sqrt(3^2·3^2·3^2)= sqrt9·sqrt9·sqrt9=3·3·3=27#

Other way would be

#9^(3/2)=(3^cancel2)^(3/(cancel 2))=3^3=27#

Answer 2 · 2018-04-13T08:10:33

#27#

#"using the "color(blue)"law of exponents"#

#•color(white)(x)a^(m/n)=(root(n)a)^m#

#rArr9^(3/2)=(root(2)9)^3=(3)^3=27#

Answer 3 · 2018-04-13T08:11:51

#27#

#9^(3/2)#

is equal to

#(9^(1/2))^3#

is equal to

#(\sqrt(9))^3#

is equal to

#3^3#

is equal to

#27#