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Exponents of exponents

Neil Geismar, TAMU
Posted 2 years ago

Mathematica tells me that a^(b c ) - (a^b)^c does equal zero, unless I assume a>0 and b>0. Is this some rule of exponents that I missed in school, have I forgotten everything important, or is this a flaw in Mathematica?

Thank you.
POSTED BY: Neil Geismar
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Ted Ersek
Ted Ersek
Posted 2 years ago

Unfortunately this is not taught well in US high schools. A related subject is that students lean that (-8.0)^(1/3) = -2.0. However, Mathematica and most software that will do that computation give a complex number. I think high schools should cover (negative)^(1/2) and (negative)^(integer) but not mention (negative)^x. I also think all high school students should lean the content in the attached.

Attachments:
Properties of Ex...pdf
POSTED BY: Ted Ersek
Neil Geismar
Neil Geismar, TAMU
Posted 2 years ago

Thank you. With sufficient assumptions, Mathematica and I now agree.
POSTED BY: Neil Geismar
Neil Geismar
Neil Geismar, TAMU
Posted 2 years ago

Thank you. When i use Assumptions -> {a > 0, Element[b, Integers]}, I get a^(b c ) - (a^b)^c = 0. Both assumptions are necessary. Could you please provide a reference for this? My problem requires b to be non-integer, so I would like more details to understand what can be done.

Again, thanks.
POSTED BY: Neil Geismar
POSTED BY: Daniel Lichtblau

Or that the exponents are integers:

Simplify[a^(b c) - (a^b)^c == 0,
 Element[b | c, Integers]]
POSTED BY: Gianluca Gorni
Ted Ersek
Ted Ersek
Posted 2 years ago

Actually we only need the following:

Simplify[(a^b)^c == a^(b c), Element[c, Integers]]

That and much more is in the document I uploaded above. A slightly obscure case that I didn't include in my attachment is the following 

Simplify[(a^b)^c==a^(b c),-1<b<1]

Both of the above are true for all complex numbers except when the second argument of Simplify specifies otherwise.
POSTED BY: Ted Ersek

I myself teach that when we do (nonpositive number)^(noninteger number) we enter a minefield, where familiar rules break down. It seems that it is not widely taught.
POSTED BY: Gianluca Gorni

Can’t check right now, but I think the equality will fail if you plug in I for all three variables.

--- edit --- My example was not quite apt. Instead we can take a=-I.

In[104]:= ayepowers = {-I^(I*I), (-I^I)^I}
N[ayepowers]

Out[104]= {I, (-I^I)^I}

Out[105]= {0. + 1. I, 2.64609*10^-18 - 0.0432139 I}

--- end edit ---
POSTED BY: Daniel Lichtblau
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