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loga=((k+1)/2)logb
(k+1)/2=loga/logb
k=2loga/logb-1
wg tego wzoru wykonam obliczenia
a=0,35; b=0,02
k1= 2*log(0,35)/log(0,02)-1=-0,4633
a=0,66; b=0,03
k2= 2*log(0,66)/log(0,03)-1=-0,7630
a=0,41; b=0,04
k3= 2*log(0,41)/log(0,04)-1=-0,4460
a=0,56; b=0,05
k4= 2*log(0,56)/log(0,05)-1=-0,6129
a=0,53; b=0,06
k5= 2*log(0,53)/log(0,06)-1=-0,54867
sprawdzenie dla k5
0,06^((-0,54867+1)/2)=0,52999