I intend to use it for treble control on my pedal. does that make it a variable resistor? (sorry, i'm a noob at this) 
I also did some googling, and read somewhere that soldering a resistor across legs lugs 1 and 3 of a linear pot results to an antilog pot. Went to tayda and purchased both linear 100k and antilog 100k pots because I wasn't sure which one is needed to make an antilog 50k pot.
Do you have a schematic of your circuit? If it really needs to be a voltage divider config, here's another option you might want to try:
buy 4 pieces or more 1/8w, 82k resistors for your C100k dual gang pots, and solder one across lugs 1 to 2, and 2 to 3 for each gang. What this does is essentially a parallel resistor effect, around 164k or resistance across the pot traces. What's different is that now, the wiper lug is involved in the movement, which I imagine will prevent any weird behaviour associated with simply soldering a resistor across lugs 1 and 3.
An antilog/reverse log potentiometer at the center of its travel will measure 10% of its total value measured from lug 1 to 2, and 90% from 2 to 3. For a 100k pot, that's 10k | 90k more or less. paralleled with 82k resistors, they drop to ~9k | ~43k which, added together, makes a ~52k pot, and considering the sort of tolerances we're dealing with, it's pretty much spot on.
The side effect is that, turned all the way to one side or another, a single 82k resistor gets paralleled with the original 100k value of the pot, which results into a ~45k resistance value. Depending on how sensitive your circuit is, this may or may not be a problem at all.
From a 100k antilog pot, you can also get ~10k by using 9.1k resistors instead. At full on or off, you get ~8.3k, while in the middle position it'll be ~4.8k | ~8.3k, or ~13.1k which is within the tolerances of typical pots.
