How do you simplify #(9^(1/2) * 9^(2/3))^(1/6)#?

2 Answers
Guilherme N.
May 22, 2015

Two important law of exponentials here:

  • #(a^n)(a^m)=a^(n+m)#

  • #(a^n)^m=a^(n*m)#

#(9^(1/2)9^(2/3))^(1/6)#=#(9^(1/2+2/3))^(1/6)#=#(9^(7/6))^(1/6)#=#9^((7/6)(1/6))#=#color(green)(9^(7/36))#

George C.
May 23, 2015

#(9^(1/2)*9^(2/3))^(1/6) = (9^((1/2+2/3)))^(1/6)#

#=(9^(7/6))^(1/6) = 9^(7/6*1/6) = 9^(7/36) = 9^(1/2*7/18)#

#= (9^(1/2))^(7/18) = (sqrt(9))^(7/18) = 3^(7/18)#

The identities we use here are:

#x^a * x^b = x^(a+b)#

#x^(ab) = (x^a)^b#

#x^(1/n) = root(n)x#

Related questions
See all questions in Fractional Exponents
Impact of this question
2405 views around the world
You can reuse this answer
Creative Commons License