[Brightcolours]

Ok, acknowledged. If you look at rather wide angle lenses, then of course the argument about small-size and low-weight becomes less relevant. However, there is another point not to be forgotten: If you compare, for instance, the Olympus 12-40/2.8 vs the Zeiss/Sony 16-70/4, as you mention above, the Olympus is a f2.8, and the Zeiss/Sony is a f4. I know, depth-of-field is similar, and the total light gathered, too. But the light gathered per sensor area is anyway larger for a f2.8 than for a f4 lens, by definition. Meaning that in case the sensor is equally sensitive in both cases, it allows you shorter exposure times at the same ISO with the f2.8 lens (or lower ISO with the same exposure time). - Or do I miss something here?

"sensor area"? 

APS-C has more sensor area. It therefore will gather more light at similar focal length/aperture ratio, by definition. 

Sensors will not be "equally sensitive". As photographer you never know just how sensitive a sensor is anyway, the changing voltage and signal amplification with the ISO setting changing is not communicated to the customer in any way. 

 

ISO is therefore meaningless. Nothing at all would necessitate to shoot MFT with f2.8 and ISO 100, and then APS-C with f4 also at ISO 100. One can set the ISO setting to whatever one desires, on both cameras. If one desires for whatever reason the same exposure time, then set the ISO on the bigger sensor camera a bit higher. Right now I am too lazy to calculate the crop factor of MFT in relation to 1.5x crop APS-C, but lets call it "X".

 

If you want an similar FOV for a 25mm MTF lens on APS-C, multiply 25mm by X. You get an equivalent focal length.

If you want an similar DOF for f2.8 on MFT with APS-C, multiply f2.8 by X. You get an equivalent f/aperture ratio.

If you then also want a similar exposure duration on MFT and APS-C when using ISO 100 on MFT, multiply ISO 100 twice with X. You get an equivalent ISO setting which will result in similar exposure duration.