Well, to calculate it the mathematical way, set it up as a ratio and solve for X:
47k / 5v = x / 8v
Get rid of "/8" by doing the opposite to the left:
(47 * 8) / 5 = x
376 / 5 = x
x = 73k. That will provide a scaled pull-down resistor value for 8v instead of 5v. We can check the actual current by using Ohm's Law. Assuming the full supply voltage was being imposed across the pulldown resistor,
I = E / R I = E / R
I = 5v / 47k I = 8v / 73k
I = 0.000106A I = 0.000109A
I = 106uA I = 109uA
Close enough. Give it a try, should work just fine. Test a +/-20% variance and see what effect it has on the circuit. I doubt the value is super-critical, but one particular value may work best.
73k * 0.20 + 73k = 87.6k
73k - 73k * 0.20 = 58.4k
If a particular pull-down current is specified as best, then calculate the resistor value for the given operating voltage by:
E = I * R
R = E / I
R = 12v / 100uA
R = 120k
Possible reasons: part count, reliability, lack of concerned about that that level of leakage, avoiding complexity, default-on behavior for user convenience (from rioraxe in the question's comments).
To expand on leakage current being of little concern: check the specification of the TPS61030: 20uA (typ). Then 1uA (max) in shutdown. What will another 20uA of leakage through the pull-up do? Expanding on laptop2d's calculations: 21uA leakage from a 1000mAh battery gives 47,600 hours of "standby" time (discounting battery self-discharge). Over 5.5 years! The self-discharge of the attached secondary cell and the usage of the device are certainly of greater power-loss concern than shutdown leakage! Leaving shutdown then trades pull-up current for the converter's quiescent current.
Thus, the expected use of this board is not greatly concerned about leakage currents in comparison to the 100 to 1000+ mA loads the battery will see in normal operation (e.g. phone charging, running an rPi).
If you were using this for something other than a USB power bank, you might be concerned about battery life. However, long-lifetime battery operated devices usually don't come equipped with 4A-switch boost converters.
Note: the proposed 200k pull-up and 400k pull-down with switch circuit would not work. From the datasheet, The device is put into operation when EN is set high. It is put into a shutdown mode when EN is set to GND. This looks to be a normal logic input, so it is intended to be driven close to the rails. It is not a comparator input like the LBO pin or certain other regulators that have precision-threshold enable inputs.
Best Answer
Yes, generally it is good practise to tie all unused inputs of digital devices to eiter GND or positive supply voltage.
You can, but that would waste power; instead tie the negative input of each unused opamp to its output (which forms a voltage follower) and the positive inputs of all unused opamps to halfway between positive and negative supply voltages, using a resistor divider (or GND in the case of split supplies). E.g. like this (shown for a single opamp):
For more details about opamp termination, see Properly terminating an unused op amp and What shall we do with the unused opamp .