Working with Log

Themis Matsoukas

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Aug 30, 2010, 6:20:14 AM8/30/10

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I can use PowerExapnd to expand a Log:

PowerExpand[Log[a b]]

Log[a] + Log[b]

How can I do the opposite, i.e. combine Log[a] + Log[b] into Log[a b]?

Thanks

Themis

Sjoerd C. de Vries

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Aug 31, 2010, 4:15:47 AM8/31/10

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Assuming[{a > 0, b > 0}, Simplify[Log[a] + Log[b]]]

Cheers -- Sjoerd

Patrick Scheibe

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Aug 31, 2010, 4:16:08 AM8/31/10

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Hi,

there's no unexpand but to know that PowerExpand uses heavy assumptions
per default is useful. This should be a warning too that the results of
PowerExpand not in general true. See for instance what

PowerExpand[Log[a b], Assumptions -> Element[{a, b}, Complexes]]

gives:

2*I*Pi*Floor[1/2 - Arg[a]/(2*Pi) - Arg[b]/(2*Pi)] + Log[a] +
Log[b]

Your expansion is valid for instance for

PowerExpand[Log[a b], Assumptions -> a > 0 && b > 0]

and therefore

FullSimplify[%, Assumptions -> a > 0 && b > 0]

gives again

Log[a b]

To see where exactly the log rule works you could do maybe

Reduce[2*I*Pi*Floor[1/2 - Arg[a]/(2*Pi) - Arg[b]/(2*Pi)] == 0, {a, b}]

Cheers
Patrick

Alexei Boulbitch

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Aug 31, 2010, 4:16:51 AM8/31/10

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Hi, Themis,
one idea may be as follows. Introduce two functions for expansion and to collect logarithms:

expandLog[expr_] := Module[{rule1, rule2, a, b, x},
rule1 = Log[a_*b_] -> Log[a] + Log[b];
rule2 = Log[a_^x_] -> x*Log[a];
(expr /. rule1) /. rule2
];
collectLog[expr_] := Module[{rule1, rule2, a, b, x},
rule1 = Log[a_] + Log[b_] -> Log[a*b];
rule2 = x_*Log[a_] -> Log[a^x];
(expr /. rule1) /. rule2 /. rule1 /. rule2
];

Then:

Log[Sin[x]*Cos[y]] // expandLog

Log[Cos[y]] + Log[Sin[x]]

Log[a + b] + Log[c + d] // collectLog

Log[(a + b) (c + d)]

Have fun, Alexei

I can use PowerExapnd to expand a Log:

PowerExpand[Log[a b]]

Log[a] + Log[b]

How can I do the opposite, i.e. combine Log[a] + Log[b] into Log[a b]?

Thanks

Themis

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Simon

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Aug 31, 2010, 4:17:02 AM8/31/10

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Note that Log[a] + Log[b] does not always equal Log[a b], that being
said...
Just write some replacement rules
logRep = x_. Log[a_] + y_. Log[b_] :> Log[a^x b^y]

Then, e.g.
In[2]:= y Log[2]+Log[x^2]-Log[z]
Out[2]= y Log[2]+Log[x^2]-Log[z]

In[3]:= %//.logRep
Out[3]= Log[(2^y x^2)/z]

Simon

Peter Breitfeld

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Aug 31, 2010, 4:14:01 AM8/31/10

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Themis Matsoukas wrote:

PowerExpand treats its expressions assuming all variables are positive.
Normally Log[ab]==Log[a]+Log[b] is not always tru, e.g.

0=Log[(-1)(-1)]!=Log[-1]+Log[-1]==2Pi I

So you have to declare the variables as positives (if your domains are right):

In[1]:= Simplify[Log[a]+Log[b], Assumptions->a>0&&b>0]
Out[1]= Log[ab]

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Bob Hanlon

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Aug 31, 2010, 4:14:12 AM8/31/10

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PowerExpand[Log[a b c]]

log(a)+log(b)+log(c)

% //. Log[x_] + Log[y_] -> Log[x*y]

log(a b c)

Bob Hanlon

---- Themis Matsoukas <tmats...@me.com> wrote:

=============

Peter

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Aug 31, 2010, 4:27:27 AM8/31/10

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try:

In[1]:= FullSimplify[Log[a]+Log[b],a>0<b]
Out[1]= Log[a b]

Peter

Themis Matsoukas

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Sep 2, 2010, 2:29:51 AM9/2/10

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To summarize the responses, there are two solutions:

1: Tell mathematica that the arguments are positive numbers:

FullSimplify[2 Log[a] + 3 Log[b], Assumptions -> {a > 0, b > 0, c > 0}]

Log[a^2 b^3]

2: Use a replacement rule

2 Log[a] + 3 Log[b] //. a_ Log[b_] + c_ Log[d_] -> Log[b^a d^c]

Log[a^2 b^3]

Thanks to all.

Themis