Will this passive non inverting anti-log circuit work?
Your 2nd circuits (with the NPN and R divider) is a good approximation of a log circuit. This is because when the input opamp generates a voltage substantially above 0.6 V, the current in the 4.7k is proportional to the voltage, and so the V across the NPN is proportional to the log of that current. The 100k &amp; 10k give a multiplier effect, so in fact, your transfer function is closer to 
VOUT = K*26mVln(Iin/Is), where Iin = (VIN-0.7)/4.7k. 'Is' is hard to find directly, but if you measure (guess) VBE at 1 mA (say 0.6 V), the equation can be rewritten as
VOUT = K[26mV*ln(In) + 0.6], where 'In' is in mA. 

K is the gain from your R divider -- with 'Contour' =0, it's 1; with 'contour' = 20k, it's 3.

Note that this circuit will change over temperature --if the NPN heats up by (say) 30 deg. C, that's equivalent to about a 10 dB reduction in volume (when you go through math).