1
$\begingroup$

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

| improve this question | |
$\endgroup$
3
$\begingroup$

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]

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ (Beat me to it...) $\endgroup$ – David G. Stork Aug 28 '18 at 23:09
  • $\begingroup$ Just remember that PowerExpand results are generally wrong for negative or complex variables. $\endgroup$ – John Doty Aug 28 '18 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 '18 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 '18 at 23:35
  • $\begingroup$ @JohnDoty, good point, yes, ... the result may not be correct everywhere. $\endgroup$ – kglr Aug 28 '18 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.