Yup. Put the effort into remembering a 3 digit log for 7 instead? Or keep the same precision with ... 7 = ~sqrt(100/2) -> (2-0.3)/2 = 1.7/2 = 0.85 Log 2

Yup. Put the effort into remembering a 3 digit log for 7 instead? Or keep the same precision with ... 7 = ~sqrt(100/2) -> (2-0.3)/2 = 1.7/2 = 0.85 Log 2 is all you need? Or even ... 7 = sqrt(sqrt(2401)) = (3*8*100)^(1/4) -> 0.95/8 + 0.3*3/4 + 2/4 = 0.12 + 0.225 + 0.5 = 0.845 VERY accurate, but that's getting to be too much.