# Exponential function not simplified by FullSimplify?

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?

• 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...
– Gert
Jan 23, 2021 at 0:20
• @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.. Feb 4, 2021 at 13:57

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};

• Or manually apply Log: a^q <= 1 // ApplySides[Log, #] & // FullSimplify[#, {0 < a < 1, q > 0}] & Apr 27, 2020 at 19:45