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)]
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)]