Very nice algorithm - no square roots and 1 digit per iteration guaranteed. I converted this into python to understand it better def log10(x): """

Very nice algorithm - no square roots and 1 digit per iteration guaranteed. I converted this into python to understand it better def log10(x): """ Find 10 decimal places of log10(x) using calculator functions only """ assert x > 0, "can't log <= 0" log = 0 while x >= 10: log += 1 x /= 10 while x < 1: log -= 1 x *= 10 # x is now [1, 10) decimal = 0.1 for i in range(10): x10 = x**10 # or calculate as above so as not to use power function while x10 >= 10: log += decimal x10 /= 10 x = x10 decimal /= 10 print(f"log10(x) = {log:.10f}, x = {x:.10f}") log10(3) Which gives log10(x) = 0.4000000000, x = 5.9049000000 log10(x) = 0.4700000000, x = 5.1537752073 log10(x) = 0.4770000000, x = 1.3220708195 log10(x) = 0.4771000000, x = 1.6313501853 log10(x) = 0.4771200000, x = 1.3349714142 log10(x) = 0.4771210000, x = 1.7977101167 log10(x) = 0.4771212000, x = 3.5253044107 log10(x) = 0.4771212500, x = 2.9646009506 log10(x) = 0.4771212540, x = 5.2439970092 log10(x) = 0.4771212547, x = 1.5726220230