已知函数f(x)满足条件f(ax-1)=lg(x+2)/(x-3))(a>0)

设:t=ax-1
则：x=(t+1)/a
(x+2)/(x-3)=[(t+1)/a+2]/[(t+1)/a-3]=(t+1+2a)/(t+1-3a)
所以,f(t)=lg[(t+1+2a)/(t+1-3a)]
即：f(x)=lg[(x+1+2a)/(x+1-3a)]
a=2时,f(x)=lg(x+5)/(x-5)
定义域:(x+5)/(x-5)＞0得x＞5或x＜-5
f(-x)=lg(-x+5)/(-x-5)=lg(x-5)/(x+5)
f(x)+f(-x)=lg(x-5)/(x+5)+lg(x+5)/(x-5)=0
f(x)是在(-无穷,-5)∪(5,+无穷)上的奇函数