QUESTION IMAGE
Question
graph f(x) = 3^x and g(x) = \log_3 x in the same rectangular coordinate system. choose the correct graph below
Step1: Analyze \( f(x) = 3^x \)
The function \( f(x)=3^x \) is an exponential function with base \( 3>1 \), so it is increasing, passes through \( (0,1) \) (since \( 3^0 = 1 \)), and has a horizontal asymptote at \( y = 0 \) (as \( x
ightarrow-\infty \), \( 3^x
ightarrow0 \)).
Step2: Analyze \( g(x)=\log_3 x \)
The function \( g(x)=\log_3 x \) is a logarithmic function with base \( 3>1 \), so it is increasing, passes through \( (1,0) \) (since \( \log_3 1 = 0 \)), and has a vertical asymptote at \( x = 0 \) (as \( x
ightarrow0^+ \), \( \log_3 x
ightarrow-\infty \)). Also, exponential and logarithmic functions with the same base are inverses, so their graphs are symmetric about the line \( y = x \).
Step3: Match with Graphs
We need to find a graph where one curve is an increasing exponential (passing through \( (0,1) \)) and the other is an increasing logarithmic (passing through \( (1,0) \)), and they are symmetric about \( y = x \). Looking at the options, we check the key points: exponential at \( (0,1) \), logarithmic at \( (1,0) \), both increasing, and symmetric. The correct graph should have these features. (Assuming the first graph (top - most) has the exponential increasing through \( (0,1) \) and logarithmic increasing through \( (1,0) \) with symmetry about \( y = x \); however, since the user's image has four graphs, we identify based on the properties. The correct graph is the one where \( f(x)=3^x \) (exponential, passes (0,1), increasing) and \( g(x)=\log_3 x \) (logarithmic, passes (1,0), increasing) are plotted with symmetry about \( y = x \).)
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The graph (identify by position, e.g., if the top - most graph has the correct curves) with the exponential \( 3^x \) (passing through (0,1), increasing) and logarithmic \( \log_3 x \) (passing through (1,0), increasing) symmetric about \( y = x \). (If we assume the first graph in the list is correct, then "The Top - Most Graph")