#!/usr/bin/perl

use 5.038;

my $j;
for my $i (2..64){
    $j = 1*($i**10); # order of magnitude
    say "$i $j -> ", log($j)/log(2);  # number of bits
}

exit 2;

__END__
  
This table shows itertion number, order of magnitude, and log base 2 of that order of magnitude.
  Aproximately it shows the conversion ratio for integer numbers to number of bits not rounded.

Math Note: Any neperian or natural logarithm log() is base 'e' where e = exp(1); # 2.7 ...
The conversion from to any base N of the logarithm of a number X is the log(X) / log (N) ratio.
  Thus,
  
  sub log10 {
      my $x = shift;
      return log($x) / log(10);
  }

sub log2 {
    my $y = shift;
    return log($y) / log(2);
}

# vid. perldoc -f log