How do you express #x^(1/2) / x^(1/3)# in radical form?

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Answer 1 · 2016-09-04T09:09:59

#root(2)x/root(3)x or root(6)x#

#x^(1/2)/x^(1/3) = x^((1/2-1/3))= x^(1/6)= root(6)x#

Answer 2 · 2016-09-04T09:17:11

#root6 x#

Use the law of indices: #" "x^m/x^n = x^(m-n)#

#x^(1/2)/x^(1/3) = x^(1/2-1/3)#

=#x^((3-2)/6)#

=#x^(1/6)#

Use the law of indices #x^(p/q)= rootq x^p#

=#root6 x#