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Re: Inverse of "PowerExpand"
*To*: mathgroup at smc.vnet.net
*Subject*: [mg51659] Re: Inverse of "PowerExpand"
*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>
*Date*: Wed, 27 Oct 2004 23:44:08 -0400 (EDT)
*References*: <clnf0f$ode$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Klaus,
let's take a slightly simpler example:
In[114]:=
v = PowerExpand[Log[f*f]]
Out[114]=
2*Log[f]
1st idea
In[115]:=
Simplify[v]
Out[115]=
2*Log[f]
... same as your example.
2nd idea
In[125]:=
ComplexExpand[v]
Out[125]=
2*I*Arg[f] + Log[f^2]
... oh, looks quite good, but the imaginary part is disturbing.
So let's tell Mathematica that f is >0 (and hence real)
In[123]:=
Simplify[ComplexExpand[v], f > 0]
Out[123]=
2*Log[f]
... still not ok!
But telling it in another way (Arg[f]==0) works:
In[124]:=
Simplify[ComplexExpand[v], Arg[f] == 0]
Out[124]=
Log[f^2]
Now back to your example:
In[1]:=
w = PowerExpand[Log[a*b*b/c]]
Out[1]=
Log[a] + 2*Log[b] - Log[c]
"Inversion" can be done by using
In[18]:=
Simplify[ComplexExpand[w], Arg[a] == 0 && Arg[b] == 0 &&
Arg[c] == 0 && a > 0 && c > 0]
Out[18]=
Log[(a*b^2)/c]
But, when playing around with the conditions, trying to be logical (a>0
should mean Arg[a]==0 etc.) and other, you can discover the most funny
things. E.g. adding the condition b>0 yields
In[19]:=
Simplify[ComplexExpand[w], Arg[a] == 0 && Arg[b] == 0 &&
Arg[c] == 0 && a > 0 && c > 0 && b > 0]
Out[19]=
2*Log[b] + Log[a/c]
Noch einen schönen Abend dabei...(have a nice evening experimenting...)
Wolfgang
Klaus G wrote:
> PowerExpand[Log[a*b*b/c]]
>
> gives:
>
> Log[a] + 2 Log[b] - Log[c]
>
> Which Function will return the original "Log"?
>
> Klaus G.
>
>
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