Is the "stereo pot" (dual gang) linear or log?
Good question, I totally forgot to mention that. It's linear. The lowest output is achieved at a 50% turn.
I did some simulations. As it turns out, replacing the 10k pot with a 1k fixes the problem for the most part. My guess is when the pot is at lets say 50%, the parts of the pots is series with the 10k input resistor become part of the feedback equation, thus lowering the amplification of the whole.
Is my assumption correct?? If so, that means the pot needs to be a fair amount smaller than the input resistors. In that case, would it be better to decrease the pots value, or increase the feedback/input resistors.