For which models was 2^3>8?

[Dave Britten]

(04-04-2017 10:26 PM)Thomas Okken Wrote:  

(04-04-2017 05:51 PM)Dieter Wrote:  The explanation is simple and straightforward: y^x is calculated as e^x · ln y.
So 2³ is not evaluated as 2·2­·2 – how far should this go? Up to the 4th power? the 5th? The 10th?

How about all the way? It's really quite efficient if you use exponentiation by squaring: x^n = (x^(n/2))^2 if n is even, and x*(x^((n-1)/2))^2 if n is odd; apply recursively or iteratively. The algorithm is O(log(n)). This is how Free42 gets 2^3=8 exactly, despite not using extended precision for intermediate results.

I'm guessing they didn't want to spend the probably very limited ROM/RAM space of the old models to handle special cases of powers like that.