I may be thinking of this too simplistically, but could you not do something along the lines of:

For each type of proposed pattern,

• Specify functions Fx and Fy as the functions to apply to the x and y values, respectively.
• Given that for a straight line, (y2 - y1) = m(x2 - x1), use the equation of a line in the form ( Fy(y2) - Fy(y1) ) = m ( Fx(x2) - Fx(x1) )
• Use the first and last data values to compute a value m
• Compute the estimated values for the internal data points
• Compute the error of the estimated values
• Select proposed pattern based on the one providing the smallest error

Is this a realistic idea, or am I thinking too simply?

Hope that helps.

Select proposed pattern based on the one providing the smallest error.... Is this a realistic idea, or am I thinking too simply?

No. I don't think you are.

The first thing that popped into my mind when I read your post was: chi-square test.

Not quite sure how (or which variation) to use it yet, but I think could be a starting point. Thanks.

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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". I'm with torvalds on this

In the absence of evidence, opinion is indistinguishable from prejudice. Agile (and TDD) debunked

There are various options to measure the "goodness of fit". It is at times also highly debated what the best criterion is. You could look for example at the R^2 of a linear fit and choose the scale with the biggest one.

There are various options to measure the "goodness of fit"

The problem is that to do a goodness of fit calculation, you need two sets of data: the actual & expected.

The only two sets that make any sense (to me at least) are the pre-scaled and post-scaled sets; but the correlation between those will (should) be perfect whichever scaling method is used, since the latter is derived mathematically from the former.

I can't see what other 'expected' values you could use?

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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". I'm with torvalds on this

In the absence of evidence, opinion is indistinguishable from prejudice. Agile (and TDD) debunked

The "expected" data set is the linear fit. The following script uses linear regression and the R^2 metrics to calculate a measure of fit for your three datasets:

use strict;
use warnings;
use Statistics::LineFit;

sub fit {
  my $fit = Statistics::LineFit->new();
  $fit->setData( @_ );
  return $fit->rSquared();
}

my @data = ( [ qw( 5 5 34 44 114 169 177 184 270 339 361 364 442 511 5
+30 554 555 587 709 709 735 778 791 859 871 899 903 926 933 952 ) ],
         [ 0.5, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 
+8191, 16383, 32767, 65535, 131071, 262143, 524287, 
           1048575, 2097151, 4194303, 8388607, 16777215, 33554431, 671
+08863, 134217727, 268435455, 536870911, 1073741823 ],
         [ 1.713125e-005, 1.748086e-006, 2.101463e-006, 1.977405e-006,
+ 3.597675e-006, 3.725492e-006, 3.924736e-006, 2.902199e-006, 
           3.988645e-006, 8.210367e-006, 3.360837e-006, 5.202907e-006,
+ 7.082570e-006, 8.778026e-006, 7.079562e-005, 9.100576e-005, 
           5.258545e-005, 9.292677e-005, 1.789815e-004, 2.113948e-003,
+ 7.229146e-004, 1.428995e-003, 2.742045e-003, 5.552746e-003, 
           1.822390e-002, 2.220999e-002, 4.316067e-002, 8.876963e-002,
+ 1.751072e-001, 3.494051e-001, 7.155960e-001, 1.347822e+000 ] );

print "    linear  loglinear     loglog\n";
for my $d (@data) {
  my @x = 1..@$d;
  my @logx = map log, @x;
  my @logd = map log, @$d;
  printf "%10.2f %10.2f %10.2f\n", fit( \@x, $d), fit( \@x, \@logd), f
+it( \@logx, \@logd );
}
[download]

The result is

    linear  loglinear     loglog
      0.99       0.69       0.95
      0.26       1.00       0.86
      0.26       0.90       0.58
[download]

which shows that the first data set describes a linear relationship while the others are more of log type (the largest R^2 wins). If you have a stats package at hand (or even Excel only) you can do the same thing and visualize the results.