I asked a similar question over on Google awhile back, this is what i got.
For the monostable:
2/3 = 1-e ^(-t/RC)
-1/3 = -e ^(-t/RC)
1/3 = e ^( -t/RC)
ln(1/3) = -t/RC
(Use your scientific calculator to find the natural logrythm (ln or e^x) of 1/3)
-1.0986123 = -t/RC
t= 1.0986123 RC
Rounding off gives: 1.1RC
For the astable:
Charging: t1 = -ln(1/2)(Ra+Rb)C
(natural log of 1/2=-0.69314718)
= 0.693(Ra+Rb)C
Discharging: t2 = -ln(1/2)(Rb)C
= .693(Rb)C
Total period T = t1+t2