How to calculate voltage divider time constant?
It boils down to the capacitance of the input pins that receive the potted-down voltage. If the capacitance is 5pF and the resistors in the attenuator project an effective parallel resistance of 1000 ohm then the 3dB point of the low pass filter formed by 1000 ohms and 5 pF is: -

\$f_C = \dfrac{1}{2\pi R C}\$ = 31.8 MHz

You'd probably get away with data speeds of about 30M bps but this is not guaranteed across the board. Here's a circuit: -







simulate this circuit &ndash; Schematic created using CircuitLab

The logic level after the attenuator is \$5V\cdot\dfrac{2.7k}{1.8k+2.7k}\$ = 3 volts

The two resistors in parallel form an effective series resistance of \$\dfrac{1.8k\cdot 2.7k}{1.8k+2.7k}\$ = 1.08k ohms