Question

graph f(x) = 2^x and g(x) = \log_{2} x in the same rectangular coordinate system. choose the correct graph below.

Explanation:

Step1: Analyze \( f(x) = 2^x \)

The function \( f(x) = 2^x \) is an exponential function with base \( 2>1 \), so it is increasing. When \( x = 0 \), \( f(0)=2^0 = 1 \), so it passes through \( (0,1) \). As \( x\to -\infty \), \( 2^x\to 0 \); as \( x\to +\infty \), \( 2^x\to +\infty \).

Step2: Analyze \( g(x)=\log_2 x \)

The function \( g(x)=\log_2 x \) is a logarithmic function with base \( 2>1 \), so it is increasing. The domain is \( x>0 \). When \( x = 1 \), \( g(1)=\log_2 1 = 0 \), so it passes through \( (1,0) \). As \( x\to 0^+ \), \( \log_2 x\to -\infty \); as \( x\to +\infty \), \( \log_2 x\to +\infty \).

Step3: Match the Graphs

We need a graph where one curve (exponential) passes through \( (0,1) \), is increasing, and the other (logarithmic) passes through \( (1,0) \), is increasing, and is defined for \( x>0 \). Also, exponential and logarithmic functions with the same base are inverses, so their graphs are symmetric about the line \( y = x \).

Looking at the options, we check the key points: exponential at \( (0,1) \), logarithmic at \( (1,0) \), both increasing, and symmetric about \( y=x \). The correct graph should have the exponential curve starting near \( y=0 \) (as \( x\to -\infty \)) and rising through \( (0,1) \), and the logarithmic curve starting near \( x=0 \) (going to \( -\infty \)) and rising through \( (1,0) \), with both increasing and symmetric about \( y=x \).

Answer:

(Assuming the correct graph is the one with the exponential curve passing through (0,1) and logarithmic through (1,0), symmetric about y=x. If the options are labeled as top, second, third, fourth, we need to identify. But from the description, the correct graph should have the exponential increasing through (0,1) and logarithmic increasing through (1,0). So the correct graph is the one that matches these properties. For example, if the second graph (from top) has these features, then the answer is the second graph. But since the images are not fully clear, but based on the functions, the correct graph is the one with \( f(x)=2^x \) (passing through (0,1), increasing) and \( g(x)=\log_2 x \) (passing through (1,0), increasing, defined for x>0) and symmetric about y=x. So the correct graph is the one that shows these two curves with these properties.)