Just figured I ask before I have to buy new pots.

Cheers

Calculating the value of the resistor is simple

Yes, but you neglected to give the OP the formula, which is a little harder to figure out . It's approx 0.13 of the value of the potentiometer. So for a 10k audio taper pot you want a 1.3k resistor. And for the record, it increases the load that the pot presents to the power rails by a factor of 10 when it's fully counter-clockwise.

To me, both solutions are just linear-ish, You can find a discussion of it here: https://www.modwiggler.com/forum/viewtopic.php?t=56318. You can see that they discuss 0.12 so 0.13 is not exact, even more so since the audio taper pot isn't exactly log. I'm not claiming that linear pots are perfect, but if it were me I'd just go buy the right pots and chalk up the extra expense as tuition. Unless there was a show breathing down my neck.

in that case you can use [+ 1] into [log 2] to convert them to a linear(ish) curve .. its probably good enough for rock and roll.

for other numbers (like the arduino's 0-255) its the same concept, [+ 1] because log is weird, [log 256] which turns it into 0-1, you can add [* 255] again or work with the normalized float

in a sense and at the ranges were interested in its sort of like a fixed look up table, so the log function will either make a linear ramp into log or make a log ramp into linear

and yeah as they mention above the log (audio) taper potentiometer is actually a linear taper with an extra resistor that is then doing resistor math (see https://sound-au.com/pots.htm and https://sound-au.com/project01.htm for a better description).