Forming product of function of prime factors

Question

Given a list of numbers each containing 2 prime factors, I wish to make a list of the products of the logs of each factor. For example, given

L = {4,6,9}, I would like to form P = {Log[2]*Log[2],Log[2]*Log[3],Log[3]*Log[3]}.

FactorInteger seems like a questionable route since it would require some work to distinguish squares from numbers with two distinct prime factors.

Any suggestions? Thank you.

Warning: Trying to force Mma to display its results in the way you like them is usually difficult and doesn't pay off — Dr. belisarius, Jun 13 '13 at 16:47
@belisarius: If you are suggesting that the best way might be a short program then I can accept that. — daniel, Jun 13 '13 at 16:56
I do not see how this can be done without, in effect, factoring the number. — Daniel Lichtblau, Jun 13 '13 at 18:24

KennyColnago · Accepted Answer · 2013-06-13 18:15:56Z

1

You could try a couple of rules, but the heads of the elements in the output list will mix Times and HoldForm...

Map[FactorInteger,{4,6,9,15,25,33,49,51}] /.
   {{{p_,2}}->HoldForm[Log[p]*Log[p]],
    {{p_,_},{q_,_}}->Log[p]*Log[q]}

answered Jun 13 '13 at 18:15

KennyColnago

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$\begingroup$ Alternatively: {4, 6, 9, 15, 25, 33, 49, 51} /. n_Integer /; PrimeOmega[n] == 2 :> Times @@ Log[Flatten[ConstantArray @@@ FactorInteger[n]]] $\endgroup$ – J. M. will be back soon♦ Jun 13 '13 at 18:23
$\begingroup$ Actually this was better than what I had anticipated, thanks. $\endgroup$ – daniel Jun 13 '13 at 23:52

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