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";
[download]

...roboticus

When your only tool is a hammer, all problems look like your thumb.

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?

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

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

The start of some sanity?

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.).

log(x + y) = log(x * (1 + y/x)) = log(x) + log(1 + y/x)

y/x = exp(log(y/x)) = exp(log(y) - log(x))

log(x + y) = log(x) + log(1 + exp(log(y) - log(x)))
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And that formula can be easily calculated using regular perl scalars because log(y) - log(x) is in [-log(x), 0] and exp(log(y) - log(x)) is in (0, 1].

Yes, something like that should work. Take the Math::Libm module, and use the expm1 and log1p functions for this. I won't figure out the correct formula now.

Update 2012-06-10: though this won't help OP, future people who stumble this question by searching might want to look at the Math::BigApprox module.

Spot on! (Though you do need to know the arbitrary base B):

sub log2p5{ log( $_[0] ) / log( 2.5 ) };;

$X = log2p5( 10 ); 
$Y = log2p5( 15 ); 
$Z = log2p5( 10 + 15 ); 
print $X, $Y, $Z;;
2.51294159473206 2.95544864408182 3.51294159473206

print 2.51294159473206 + log2p5( 1 + 2.5**( 2.95544864408182 - 2.51294
+159473206 ) );;
3.51294159473206
[download]

---

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

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

The start of some sanity?

yes, but it is easy to convert from any base to natural, perform the computation and then convert back to the original base.

Thanks! That will save one exponentiation step. Earlier I had to do base^x, base^y, now only one exponentiation step is required (exp(log(y) - log(x))). Btw, range for log(y) - log(x) is (-inf 0]

As I came to learn from another source where they CREATE log table for this particular problem (function logmath_init() in http://cmusphinx.svn.sourceforge.net/viewvc/cmusphinx/trunk/sphinxbase/src/libsphinxbase/util/logmath.c?revision=11275&view=markup)

It's like this - first set (log(y) - log(x)) = 0; then compute base^(1 + exp(log(y) - log(x))

then decrement log(y) - log(x) by 1 (equivalent to dividing by base) and recomputing base^(1 + exp(log(y) - log(x))

this is done untill we reach the MOST negative number. Using this table one can back-compute any log(x+y) for that particular base.

Thanks again for your help.