2^(10/3) ~= 10

so

    2^(100/30) ~= 100

so it should work.

And I think this trick should work for powers of any a in any base b where there is a k such that

    a^(b/k) ~= b 

aka

    k ~= b * log a / log b
    a ~= e^ (k * (log_e b)/ b)

and if b is the base of your logarithm, this simplifies to

    k ~= b * log a
    a = b^(k/b)

   10 * log 2 = 3.0103
    
   10 * log 5 = 6.9897 

, so 5 works too....

    >>> approx = lambda a: sorted( [float(scale(str(a**k)[:3])) for k in range(0,10)])

    >>> [10**(k/10.0) for k in range(0,10)]

    [1.0, 1.25, 1.58, 1.99, 2.51, 3.16, 3.98, 5.01, 6.30, 7.94]

    >>> approx(2)
    [1.0, 1.28, 1.60, 2.00, 2.56, 3.20, 4.00, 5.12, 6.40, 8.00]

    >>> approx(5)
    [1.0, 1.25, 1.56, 1.95, 2.50, 3.12, 3.90, 5.00, 6.25, 7.81]

Is it a coincidence that this works for 2 and 5 in base 10, while 2*5 = 10 ?