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I'm trying to understand why the first expression doesn't simplify while the second does:

FullSimplify[a^q <= 1, {0 < a < 1, q > 0}]
FullSimplify[q Log[a] <= 0, {0 < a < 1, q > 0}]

Aren't these expressions equivalent under the supplied assumptions?

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  • $\begingroup$ Whether the second form is simpler than the first is more like a matter of opinion I'm afraid. I see no reason why Simplify would make any changes in the first place... $\endgroup$
    – Gert
    Jan 23, 2021 at 0:20
  • $\begingroup$ @Gert I'm not claiming the second function as written is simpler, but both should evaluate to "True" because they are both always satisfied. The second does, but the first remains what it is.. $\endgroup$
    – mimuller
    Feb 4, 2021 at 13:57

1 Answer 1

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I ended up using this workaround with Reduce instead. Not super pretty, but at least it shows that the algebra is correct. Applied to this example,

assum = {0 < a < 1, q > 0};
FullSimplify[Reduce[And@@ assum && a^q <= 1], assum]

I will wait a bit before accepting this answer in case anyone else has a more elegant approach, but didn't want you to go through the pain of typing out something that already has an answer elsewhere.

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    $\begingroup$ Or manually apply Log: a^q <= 1 // ApplySides[Log, #] & // FullSimplify[#, {0 < a < 1, q > 0}] & $\endgroup$
    – Bob Hanlon
    Apr 27, 2020 at 19:45

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