in reply to Add numbers in log domain [solved]

pRaNaV:

You'll likely have to clarify your question. As I read it, you're probably wanting to compute the logarithm of a sum using a base other than e. But since the documentation for the log function (which you can read via perldoc -f log) covers this, I'm hard pressed to imagine that's what you actually want. But if you did want that, then:

my ($x, $y, $base) = (23.5, 54.3, 2.0);
my $result = log($x + $y) / log($base);
print "result: $result\n";

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Re^2: Add numbers in log domain
by BrowserUk (Patriarch) on Jun 04, 2012 at 11:32 UTC

    My interpretation of the OP is: Given two (log) values X & Y where both are the logs to the same (unspecified) base: B; is it possible to derive Z, such that Z = logB( B^X + B^Y ); without going through the exponentiation step.

    Eg.

    X = logB( x ) = 2.51294159473206;

    Y = logB( y ) = 2.95544864408182;

    Z = logB( x + y ) = 3.51294159473206;

    Is it possible to derive Z without discovering x & y by 'antilogging' X & Y?

    With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
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[reply]

      Sorry, for not being clear. Yes, BrowserUK is right. I wan't to know how this (log(x + y)) can be done in perl without using exponentiation (base^x, base^y etc.).

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