which of the following shows the graph of $y = 2\\ln x$?

Explanation:

Step1: Analyze the function ( y = 2\ln x )

The domain of ( y=\ln x ) is ( x>0 ), so ( y = 2\ln x ) is also defined for ( x>0 ). At ( x = 1 ), ( y=2\ln(1)=0 ) (since ( \ln(1) = 0 )). The function ( y = \ln x ) passes through ( (1,0) ) and is increasing, and ( y = 2\ln x ) is a vertical stretch of ( y=\ln x ), still increasing, defined for ( x>0 ), passing through ( (1,0) ).

Step2: Analyze the graphs

The first graph (top) has a part where ( x<0 ), but ( y = 2\ln x ) is not defined for ( x\leq0 ), so it's incorrect. Wait, maybe the second graph (bottom) is the correct one? Wait, let's re - check. Wait, the function ( y = 2\ln x ): when ( x = 1 ), ( y = 0 ); as ( x\rightarrow0^{+} ), ( y\rightarrow-\infty ); as ( x\rightarrow+\infty ), ( y\rightarrow+\infty ), and it's increasing. The first graph (top) has a point at ( x = 0 ) and ( x<0 ), which is invalid. The second graph (bottom) is defined for ( x>0 ), starts from below (as ( x\rightarrow0^{+} ), ( y\rightarrow-\infty )) and increases, passing through ( x = 1 ) (where ( y = 0 ))? Wait, maybe the second graph (the lower one) is the correct one. Wait, the first graph (top) has a curve that goes through ( x = 0 ), but ( \ln x ) is not defined at ( x = 0 ). So the correct graph should be the one where the curve is only for ( x>0 ), starts from the bottom (as ( x\rightarrow0^{+} ), ( y\rightarrow-\infty )) and increases, passing through ( (1,0) ). So the second graph (the lower one) is the correct one. But since the user's image shows two graphs, the correct one is the lower graph (the one with ( y ) - axis and ( x>0 ) curve). Wait, maybe the first graph in the image (the top one) has an error in the ( x<0 ) part, so the correct graph is the second one (the lower one).

Answer:

The lower graph (the one below the first graph, the second graph in the image)