I wrote: > So with just 2× = 3.01dB we already know (...) 8, to fairly good precision. (...) (The most imprecise of these is 8: 10⁰·⁹ is about 7.94, which is a

I wrote: > So with just 2× = 3.01dB we already know (...) 8, to fairly good precision. (...) (The most imprecise of these is 8: 10⁰·⁹ is about 7.94, which is an error of about -0.7% when the right answer was 8.) This is slightly mixed up. It's true that approximating 8× as 9dB gives you an 0.7% error. But I was talking about 3.01dB, and approximating 8× as 9.03dB gives 7.998×, only about 0.02% low. The errors for the logarithms of the 10 digits thus approximated are: >>> print('\n'.join([f' {10*p:.2f}: {100*(10**p-i-1)/(i+1):+.3f}%' for i, p in enumerate([0, .301, .477, .602, .699, .778, .845, .903, .954])])) 0.00: +0.000% 3.01: -0.007% 4.77: -0.028% 6.02: -0.014% 6.99: +0.007% 7.78: -0.035% 8.45: -0.023% 9.03: -0.021% 9.54: -0.056%