Re: natural log of E

• To: mathgroup at christensen.cybernetics.net
• Subject: [mg2009] Re: natural log of E
• From: "Brian J. Albright" <albright at physics.ucla.edu>
• Date: Thu, 14 Sep 1995 23:27:46 -0400
• Organization: UCLA Department of Physics and Astronomy

```"Paul C. Hines" <hines at bonnie.drea.dnd.ca> wrote:
>I'm trying to do some operations on the exponent of E.
> I've tried to use the form test = Log[E[n]] so as to
>return the value, test = n and thereby work with the exponent
> n. Instead Mathematica returns Log[E[n]] to me.  Why is that
> and how do I get around the problem?
>
>
>

Well, at the risk of sounding trite, I'd first have to suggest
that you try "Log[Exp[n]]" rather than "Log[E[n]]"  That way
instead of getting "Log[E[n]]" back, you get back "Log[E^n]".

Big improvement, eh?

I suspect that the reason that Mathematica is not able to
interpret "Log[E^n]" as "n" might have to do with the fact that
"Log" is a multivalued function of its argument, and Mathematica
isn't able to decide on which Riemann sheet to put the answer.
However, this isn't an entirely satisfactory answer, since when I
ask for "Log[E]", mm spits out "1" rather than "1 + 2 Pi I n_Integer"
or somesuch.  /*shrug*/  Beats me.  I guess that's a question for
the experts.

You can force the issue by applying a rule that handles Log[E^n].
This really isn't satisfactory either, but it can work in a pinch if
you know what Riemann sheet you want: e.g.

In[1]:= BogusLogRules = {Log[E^n_] -> n, Log[ m_ E^n_] -> Log[m] + n};

In[2]:= expression = Log[  2 Exp[4 x] Log[ 3 Exp[5x] ]  ];

In[3]:= expression

4 x        5 x
Out[3]:= Log[2 E    Log[3 E   ]]

In[4]:= expression //. BogusLogRules

Out[4]:= 4 x + Log[2 (5 x + Log[3])]

There are undoubtedly cleaner/more clever ways to do this, but you
get the point....  Just be careful about how you handle the
multivaluedness.

Regards.

-Brian

--
Brian J. Albright
Department of Physics and Astronomy, UCLA
albright at physics.ucla.edu
http://bohm.physics.ucla.edu/~albright

```

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