已知y=y1+y2,其中y1与log(3)x成正比,y2与log(3)x成反比,且当x=1/9时,y1=2;当x=1/27时,y2=-3,试确定函

Question

已知y=y1+y2,其中y1与log(3)x成正比,y2与log(3)x成反比,且当x=1/9时,y1=2;当x=1/27时,y2=-3,试确定函

Answer 1

∵ y1与log(3)x成正比,y2与log(3)x成反比
设y1=k1×log(3)x, y2=k2÷log(3)x
当x=1/9时,y1=2;当x=1/27时,y2=-3,代入上面两个式子
得 y1 = k1×log(3)(1/9) = -2×k1 = 2
y2 = k2÷log(3)(1/27) = k2÷(-3) = -3
∴ k1=-1, k2=9
∴ y1= -log(3)x
y2= 9 /[log(3)x]
∴ y= -log(3)x + 9 /[log(3)x]