Re: an expression with logs

*To*: mathgroup at smc.vnet.net*Subject*: [mg75385] Re: an expression with logs*From*: Norbert Marxer <marxer at mec.li>*Date*: Fri, 27 Apr 2007 05:29:13 -0400 (EDT)*References*: <f0n78d$qdg$1@smc.vnet.net>

On 25 Apr., 11:33, dimitris <dimmec... at yahoo.com> wrote: > Hello. > > In[27]:= > oo = 5*Log[4/3] - 3*Log[3/2] - Log[2]; > > I want oo to take the form 4*(3*Log[2] - 2*Log[3]) > > I can think one workaround like > > In[43]:= > PowerExpand[oo] > Collect[%, {Log[2], Log[3]}] > % /. Log[4] -> 2*Log[2] > Factor[%] > > Out[43]= > -Log[2] - 3*(-Log[2] + Log[3]) + 5*(-Log[3] + Log[4]) > > Out[44]= > 2*Log[2] - 8*Log[3] + 5*Log[4] > > Out[45]= > 12*Log[2] - 8*Log[3] > > Out[46]= > 4*(3*Log[2] - 2*Log[3]) > > But I am sure there is something smarter and more compact. > > Any ideas? > > Dimitris Hello This version should work on arbitrary rational and integer numbers. (Note that I didn't test it thoroughly though). f1 := #1 /. Log[Rational[x_, y_]] :> Log[x] - Log[y] & f2 := #1 /. Log[x_Integer] :> Plus @@ f3 /@ FactorInteger[x] & f3[{x_, y_}] := y*Log[x]; f := Expand[f2[f1[#1]]] & Then f[Log[24]] f[5*Log[4/3] - 3*Log[3/2] - Log[2]] f[Log[(3*4^3)/5^2]] will give you 3*Log[2] + Log[3] 12*Log[2] - 8*Log[3] 6*Log[2] + Log[3] - 2*Log[5] Best Regards Norbert Marxer