Electronic – Resistance of a Wheatstone Bridge

Wheatstone bridges are mostly used for very small resistance changes in a sensor, like for a strain gage or something like that. Your resistance change is huge, so you don't need one. As most have been saying, put a constant current source into your sensor. In the attached schematic, as the noninverting input of the op amp is at virtual ground, the current going through the sensor will be a constant 1ma. You haven't offered enough of a response profile to allow us to help you to linearize (or if you even need to), but log and inverse log amps can be found here. I have a feeling an inverse log amp will help. Alternatively, you can sample the output (after inverting it once to make it positive), and use a lookup table if that gives you enough sensitivity in your range of interest

^{simulate this circuit – Schematic created using CircuitLab}

The whole point is to match 2 different resistor values to each other, correct?

In that case, get 2 high quality fairly closely matched resistors R1, R3.

Then irregardless of the actual values of R2 and Rx, as long as they are closely matched the voltage divider circuits R1/R2 and R3/Rx will produce a near-zero voltage Vg. The smaller the voltage, the closer the match (assuming very good R1/R3 matching). You can now use R2/Rx in your circuits and they will be pretty well matched to each other.

There is another issue with resistor matching, though: temperature coefficients. These are often specified as max values and there's no telling if two given resistors will drift at the same rate or even in the same direction.