Time taken to charge a capacitor
Here are the schematic and naming conventions used :



simulate this circuit &ndash; Schematic created using CircuitLab
Charge and discharge bounds
You probably know that already, but the theory of operation is as follows :
When TRIGGER is pulled at \$2/3 \cdot V_{CC}\$, the output goes \$\text{LOW}\$, and the DISCHARGE pin becomes a short to ground.
\$C_1\$ is then discharged through \$R_2\$ to ground.
When the THRESHOLD's pin voltage (and \$C_1\$'s voltage reaches \$1/3\cdot V_{CC}\$, the output goes \$\text{HIGH}\$, the DISCHARGE pin becomes open-circuited, and \$C_1\$ charges to \$V_{CC}\$ through \$R_1\$ and \$R_2\$.
The \$\text{HIGH}\$ period is then the time the capacitor takes to charge from \$1/3\$ to \$2/3\$ of the supply through the equivalent resistance \$R_{tot} = R_1 + R_2\$.
The \$\text{LOW}\$ period is the time the capacitor takes to discharge from \$2/3\$ to \$1/3\$ of the supply through \$R_2\$.
The series RC circuit charge-discharge formula
The charge-discharge formula of an RC circuit is :
$$V_C(t) = \Delta V(1 - e^{-t/RC}) + V_I$$
Where

\$V_I\$ is the initial voltage across the capacitor.
\$\Delta V\$ is the difference between the initial voltage \$V_I\$ and the steady state voltage once the circuit will have stabilized at the limit.

Often, it is assumed \$V_I = 0\$ and the formula takes the more well-known form of :
$$ V_C(t) = V_{\text{terminal}}(1 - e^{-t/RC}) $$
Application to the 555 timer's astable mode circuit
Only the charging part will be discussed here, the discharging part of the period follows the same principles.
During the charging part of the cycle, the capacitor charges (if the 555 didn't exist) to \$V_{CC}\$ from \$1/3 \cdot V_{CC}\$. It is trivial to see that , \$V_I = 1/3\cdot V_{CC}\$ and \$\Delta V = 2/3 \cdot V_{CC}\$.
For \$t = 0\$ the moment the capacitor begins to charge, the \$\text{HIGH}\$ period is \$t\$ such that \$V_C(t) = 2/3 \cdot V_{CC}\$, as when this is satisfied, the capacitor begins its discharge, and the output becomes \$\text{LOW}\$.
$$
\begin{align}
2/3 \cdot V_{CC}(1 - e^{-t/R_{tot}C}) + 1/3\cdot V_{CC} = 2/3\cdot V_{CC}
&amp; \Rightarrow 2/3(1 - e^{-t/R_{tot}C}) + 1/3 = 2/3 \\
&amp; \Rightarrow 1 - e^{-t/R_{tot}C} + 1/3 \cdot 3/2 = 1 \\
&amp; \Rightarrow -e^{-t/R_{tot}C} = 1 - 1 - 1/2 = -1/2 \\
&amp; \Rightarrow -t/(R_{tot}C) = \ln(1/2) \\
&amp; \Rightarrow t = -\ln(1/2) \cdot (R_{tot}C) \approx 0.693 \cdot R_{tot}C
\end{align}
$$
Changing what needs to be, the discharge part of the cycle follows the same line of reasoning.