On 3/11/06, Robert Dodier <robert.dodier at gmail.com> wrote:
> Bear in mind that the expand_log10 function (already defined
> in log10.mac) replaces instances of log10(x) with log(x)/log(10),
> so if some operator FOO doesn't know what to do with log10,
> try applying expand_log10 and then FOO.
Is there then a contract_log10 that will take the answer, which will
be in terms of log, and put it into terms of log10?
--Joel