由 Raceleader 於 星期五 四月 04, 2003 7:43 pm
This is only my guess
a,b,k≠0
After testing, k≠1/3, 1/2, 1
If 0<k<1/3
logk=-(ak+b) ---(1)
log2k=-(2ak+b) ---(2)
log3k=-(3ak+b) ---(3)
(1)-(2):
log(1/2)=ak
(2)+(3):
log(2/3)=ak
Which is impossible
If 1/3<k<1/2
logk=-(ak+b) ---(1)
log2k=-(2ak+b) ---(2)
log3k=3ak+b ---(3)
(1)-(2):
log(1/2)=ak
(2)+(3):
log(6k2)=ak
6k2=1/2
k=1/√12 (Rejected)
If 1/2<k<1
logk=-(ak+b) ---(1)
log2k=2ak+b ---(2)
log3k=3ak+b ---(3)
(1)+(2):
log(2k2)=ak
(3)-(2):
log(3/2)=ak
2k2=3/2
k=√3/2
If 1<k
logk=ak+b ---(1)
log2k=2ak+b ---(2)
log3k=3ak+b ---(3)
(2)-(1):
log2=ak
(3)-(2):
log(3/2)=ak
which is impossible
k=√3/2
After checking a, b value, it does not fix value