How do you simplify #root5(x^3)/root7(x^4)#?

2 Answers
Jul 30, 2016

Answer:

#" "x^(3/5)/x^(4/7) = x^(1/35) = root(35)(x) larr" 35th root"#

Explanation:

Write as: #(x^(3/5))/(x^(4/7)#

This is the same as: #x^(3/5-4/7)#

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Consider the #3/5-4/7#

Write as #(3/5xx1)-(4/7xx1)#

This is the same as:#(3/5xx7/7)-(4/7xx5/5)#

#=21/35-20/35 = (21-20)/35= 1/35#
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So #" "x^(3/5)/x^(4/7) = x^(1/35) = root(35)(x) larr" 35th root"#

Jul 30, 2016

Answer:

# x^(1/20) = root20(x)#

Explanation:

If both roots were the same we could have combined them into the root of a single fraction. But they are different.

Change to index form using : #" "rootq(x^p) = x^(p/q)#

#root5(x^3)/root7(x^4) = x^(3/5)/x^(4/7) " simplify using" x^m/x^n = x^(m-n)#

=#x^(3/5-4/7)#

=#x^((21-20)/35)#

=# x^(1/35) = root35(x)#