(i)看上面
(ii)

h(x)=ln[g(e^x)]
h(x)=ln[g(e)g(e)g(e)...g(e)]  <-共x個g(e)
h(x)=ln{[g(e)]^x}
h(x)=ln(2^x)

so  ln[g(e^x)]=ln(2^x),
i.e.    g(e^x)=2^x

g(e^x)=2^x
g(e^x)=2^(xlne)    because of lne=1
g(e^x)=2^[ln(e^x)]
g(x)=2^lnx

(i)
put y=1
g(x)=g(x)g(1)
g(x)[1-g(1)]=0
suppose g(1) 不等於1
g(x)=0  which is contradiction(因為g is a non constant function)
so g(x)不等於0

hence
你put y=x 再用contradiction 試下就計到架啦

(ii)未諗到~