When letting Mathematica simplify

(a^(-5) b (a^2 c^3)^3)/((a b^2 )^(-2) c^7)

it returns the 'correct' answer

a^3 b^5 c^2

But, when entering

((a^5 b)^(1/2) c^3)/(a^5 (b^4 c^5)^(1/4))

the return is far from as 'correct' as the above. (Can it be the 1/2 and 1/4?)

How can I get Mathematica to always return an answer with

a^(...) b^(...) c^(...)

even if that means (negative) fractions in (...)? And, can negative exponentials be ruled out so that Mathematica responds with a fraction with expressions in numerator and denominator with only positive exponentials? TIA


1 Answer 1


You can use PowerExpand (which implicitly assumes positive real values for variables raised to fractional powers):

 PowerExpand[((a^5 b)^(1/2) c^3)/(a^5 (b^4 c^5)^(1/4)) ]

c^(7/4)/(a^(5/2) Sqrt[b])


$\frac {c^{7/4}} {a^{5/2}\sqrt {b}}$

Alternatively, Refine or FullSimplify with Assumptions -> {a > 0, b > 0, c > 0}:

Refine[((a^5 b)^(1/2) c^3)/(a^5 (b^4 c^5)^(1/4)),  Assumptions -> {a > 0, b > 0, c > 0}]

c^(7/4)/(a^(5/2) Sqrt[b])

FullSimplify[((a^5 b)^(1/2) c^3)/(a^5 (b^4 c^5)^(1/4)), 
 Assumptions -> {a > 0, b > 0, c > 0}]

c^(7/4)/Sqrt[a^5 b]

  • 1
    $\begingroup$ (Beat me to it...) $\endgroup$ Aug 28, 2018 at 23:09
  • $\begingroup$ Just remember that PowerExpand results are generally wrong for negative or complex variables. $\endgroup$
    – John Doty
    Aug 28, 2018 at 23:15
  • $\begingroup$ Is there some way to avoid square root signs and just return "powers", even if the exponential is a fraction? $\endgroup$
    – mf67
    Aug 28, 2018 at 23:20
  • $\begingroup$ @mf67, you can use PowerExpand[((a^5 b)^(1/2) c^3)/(a^5 (b^4 c^5)^(1/4))], p : Power[_, Rational[-1 | 1, 2]] :>Inactivate[p, Power] $\endgroup$
    – kglr
    Aug 28, 2018 at 23:35
  • $\begingroup$ @JohnDoty, good point, yes, ... the result may not be correct everywhere. $\endgroup$
    – kglr
    Aug 28, 2018 at 23:47

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