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MathGroup Archive 1999

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Re: Simplify Log[ab] - Log[b] to Log[a] ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg16071] Re: [mg16039] Simplify Log[ab] - Log[b] to Log[a] ?
  • From: dreiss at !SPAMscientificarts.com (David Reiss)
  • Date: Tue, 23 Feb 1999 03:45:24 -0500
  • Organization: EarthLink Network, Inc.
  • References: <199902210515.AAA28765@smc.vnet.net.> <7aquud$387@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <7aquud$387 at smc.vnet.net>, Jurgen Tischer
<jtischer at col2.telecom.com.co> wrote:

> In[1]:= PowerExpand[Log[a b] - Log[b]]
> 
> Out[1]= Log[a]
> 
> Jurgen
> 
> Simon Allfrey wrote:
> > 
> > How do I persuade Mathematica to simplify
> > 
> > Log[a b] - Log[b] to Log[a]?
> > 
> > when processing algebraic expressions?


I missed the idea of using PowerExpand in my posting
so I have (re)learned something here.  
I guess the thing worth pointing  out for the user 
is the usual spiel about how  PowerExpand doesn't 
respect branch cuts etc... so that using 
it on an expression that contains something like
Log[a b] - Log[b] can lead to unwanted consequences
beyond the (presumably desired) consequences of 
changing Log[a b] - Log[b]  to Log[a].  For example in 

In[1]:= (z^2)^(1/2) (Log[a b]-Log[b])//PowerExpand

Out[1]= z Log[a]


this might not be what the user wants (if z<0 for example).
However,


In[3]:=
(z^2)^(1/2) (Log[a b]-Log[b])/.{Log[x_ y_]:>Log[x]+Log[y]}

Out[3]= Sqrt[z^2]*Log[a]

It all depends on the broader context of the problem.

Regards,

David

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