设f(xx-1)=logxx/xx-2且f(g(x))=logx,求g(x)

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设f(xx-1)=logxx/xx-2且f(g(x))=logx,求g(x)

设f(xx-1)=logxx/xx-2且f(g(x))=logx,求g(x)
设f(xx-1)=logxx/xx-2且f(g(x))=logx,求g(x)

设f(xx-1)=logxx/xx-2且f(g(x))=logx,求g(x)
f(xx-1)=logxx/xx-2
请问你这xx是啥意思?
x的平方?
后面的logxx/xx-2是log(x^2/(x^2-2))?
令y=x^2/(x^2-2) 则
x^2=2y/(y-1)
x^2-1=(y+1)/(y-1)
所以 f((y+1)/(y-1))=log(y)
g(y)=(y+1)/(y-1)
即g(x)=(x+1)/(x-1)

f(x^2 -1)=log(x^2/(x^2-2)) => x^2/(x^2-2)>0 =>x^2-2 >0=>x^2 >2
=>x^2-1>1 ;let x^2-1=t => f(t)=log((t+1)/(t-1) ;t>1
=>f(g(x))=logx=log((g(x)+1)/(g(x)-1)) ; x>0
=>x=(g(x)+1)/(g(x)-1)
=> x*g(x)+x =g(x)+1 => (x-1)*g(x)=1-x
=>g(x)=-1 ;x>√2