How do you write #(-28)^(7/5)# in radical form?

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Answer 1 · 2017-06-06T20:00:34

#(root(5)(-28))^7#

#m^(1/n)-=root(n)(m)#

#(-28)^(5/7)#

#=((-28)^(1/5))^7#

#=(root(5)(-28))^7#

Answer 2 · 2017-06-06T20:04:07

See a solution process below:

We can rewrite this expression as:

#(-28)^(7 xx 1/5)#

We can now use this rule of exponents to rewrite the expression again:

#x^(color(red)(a) xx color(blue)(b)) = (x^color(red)(a))^color(blue)(b)#

#(-28)^(color(red)(7) xx color(blue)(1/5)) = (-28^(color(red)(7)))^color(blue)(1/5)#

We can now use this rule for exponents and radicals to put the expression into radical form:

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

#(-28^7)^(1/color(red)(5)) = root(color(red)(5))(-28^7)#