Electronic – How is the positive input of this op-amp calculated

You have some boundary conditions: In your circuit the collector voltage is zero because the inverting input is a virtual ground. Also in the transdiode configuration, Ic=Ie, and Ve=Vout.

Use a reasonable approximation to the Ebers-Moll equation for currents that are greater than a uA. $$I_{c}\; =\; I_{s}\; e^{-\frac{qV_{E}}{kT}}$$ Is is emitter reverse saturation current. You solve for a variable in the exponent of an exponential by using its inverse, the natural log. $$\ln \left( \frac{I_{c}}{I_{s}} \right)\; =\; \frac{qV_{E}}{kT}$$ and solve for V $$V_{E}\; =\; -\frac{kT}{q}\; \ln \left( \frac{I_{c}}{I_{s}} \right)$$ This is V across the transistor junction and because the virtual ground at the inverting input is zero, this is V on the output, and is negative (inverted of course).

In your circuit, current into the virtual ground must equal current out, so current through the transistor is the same as current through the resistor. This means $$I_{c}\; =\; \frac{V_{in}}{R_{1}}$$ and substituting $$V_{E}\; =\; V_{o\; }=\; -\frac{kT}{q}\; \ln \left( \frac{V_{in}}{I_{s}R_{1}} \right)$$ kT/q has units of volts so for simplicity, replace with a scaling voltage in front. $$V_{o\; }=\; -V_{t}\; \ln \left( \frac{V_{in}}{I_{s}R_{1}} \right)$$

This is the first time I have used LaTex so I was a little distracted. Hope it makes sense. There are approximations and simplifications I can expand on.

The op-amp can't do anything useful in this configuration. You need to give it more supply voltage than the desired output voltage, even though it's "rail to rail".

If you ask it to output 8V or even 11V with a 12V supply, it may be able to do that, provided the load is not too much for the op-amp to supply. In this case, that might be 10mA or so.

I don't know exactly why you're only getting 4V at the output, but I suspect it's because the load is something like a couple hundred ohms or equivalent, so the output current is limiting at 20-30mA.

Even if you put a lighter load on there, my first comment applies- this circuit can not do anything useful without some changes (more supply voltage compared to desired output voltage).