g(x)=e^x,x0则g(g(1/3))=

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g(x)=e^x,x0则g(g(1/3))=g(x)=e^x,x0则g(g(1/3))=g(x)=e^x,x0则g(g(1/3))=1/3>0所以g(1/3)=ln(1/3)=-ln3所以g[g(1/

g(x)=e^x,x0则g(g(1/3))=
g(x)=e^x,x<=0 g(x)=Inx,x>0则g(g(1/3))=

g(x)=e^x,x0则g(g(1/3))=
1/3>0
所以g(1/3)=ln(1/3)=-ln3<0
所以g[g(1/3)]=e^]g(1/3)]
=e^[ln(1/3)]
=1/3