唔識做一條數

maxycy

(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