# Simplifying algebraic expressions with fractions in exponential

When letting Mathematica simplify

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


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

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])

 TeXForm[%]


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

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]

• (Beat me to it...) – David G. Stork Aug 28 '18 at 23:09
• Just remember that PowerExpand results are generally wrong for negative or complex variables. – John Doty Aug 28 '18 at 23:15
• Is there some way to avoid square root signs and just return "powers", even if the exponential is a fraction? – mf67 Aug 28 '18 at 23:20
• @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] – kglr Aug 28 '18 at 23:35
• @JohnDoty, good point, yes, ... the result may not be correct everywhere. – kglr Aug 28 '18 at 23:47