Same LED, different voltage drops with different supply voltages
I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf =  ≈ 7 mA
3.3V with a 325 Ω resistor and Vf =  ≈ 2 mA 

  

plugging your numbers in to a resistor calculator:

  


Source: LED Series Resistor Calculator  





Looking on an IV curve:  

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V  


Source: OSRAM blue LED  




  is there any mathematicval formula to calculate voltage drops at
  different curerrent ? – Anton Stafeyev   
  
  it is very important to know exact forward voltage so i can se the
  luminosity as exact as possible. how would one do it ? – Anton
  Stafeyev


This is easier to show with a high power LED.   

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED.
This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.     

  



We have to up the current from 400 ma to get 1000 lumens. 

1000/900 = 111%  

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give us 1000 lumens.  





We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.  

The forward voltage for this LED at 475 mA is 17.75V. 





Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.   

So for 1000 lumens we need a 13.3Ω, 5 Watt  resistor.