You are correct. If the collector current is different in each transistor then VBE is different in each of the transistors.
- But the difference is small enough that it doesn't substantially affect the answer so it was neglected.
- Additionally, making the assumption VBE1 = VBE2 keeps all the circuit equations linear rather than equations involving exponentials. That makes the math much easier.
The base-emitter voltage difference in the two transistors will vary logarithmicaly with their ratio, so you will get very small changes in Vbe even with fairly substantial changes in emitter current.
From the basic equations for the BJT...
Ic = Is * (e^(Vbe/Vt) - 1) ≈ Is * e^(Vbe/Vt)
Vbe ≈ Vt * ln(Ic / Is)
For two matched (Is1 = Is2 = Is) transistors Q1 and Q2...
Vbe1 - Vbe2 ≈ Vt * ln(Ic1 / Is) - Vt * ln(Ic2 / Is)
Vbe1 - Vbe2 ≈ Vt * ln(Ic1 / Ic2)
From the form of this equation we see that the difference in base emitter voltage varies with the logarithm of the ratio of the collector currents. Vt is usually small (like 26mV at room temperature).
So having IC2 = 2.718 x IC1 only results in a 26mV difference in base emitter voltage.
If your circuit was biased so that there was substantially more than 26mV across R2 (like say 1V), then making the assumption that VBE1 = VBE2 only leads to about 2~3% error in the calculation compared to a more detailed analysis.
But in the real world even matched transistor pairs are only matched to within a few percent anyways, so making a calculation more accurate than that usually serves little purpose.