第1个回答 2012-02-04
A(1,0),B(0,1),AB:y=1-x
∵BM = MN = NA
∴xM=1/3, yM=2/3
xN=2/3,yN=1/3
将M(1/3,2/3)代入y=x^a,
得:2/3=(1/3)^a==>a=log(底1/3)(2/3)
将N(2/3,1/3)代入y=x^b
得:1/3=(2/3)^b==>b= log(底2/3)(1/3)
∴a=log(底1/3)(2/3) b= log(底2/3)(1/3)
第2个回答 2012-02-04
A(1,0),B(0,1),
∵M,N是三等分点,BM = MN = NA
∴M(1/3,2/3),N(2/3,1/3)
即y=x^a过M(1/3,2/3), ==> 2/3=(1/3)^a ==>a=log(1/3)2/3=1-log3(2)
y=x^b过N(2/3,1/3), ==> 1/3=(2/3)^b ==>b=log(2/3)1/3=1-log(2/3)(2)