score on last try: 0 of 1 pts. see details for more. you can retry this question below match each equation…

Explanation:

Step1: Analyze ( e^x ) and ( 10^x )

Exponential functions ( y = a^x ) ( ( a>1 )) increase, and ( 10^x ) grows faster than ( e^x ) (since ( 10 > e )). The green (G) curve is steeper than blue (B), so ( 10^x ) matches green (G) and ( e^x ) matches blue (B).

Step2: Analyze ( \ln(x) ) and ( \log(x) )

Logarithmic functions ( y = \log_a(x) ) ( ( a>1 )) increase, and ( \ln(x)=\log_e(x) ), ( \log(x)=\log_{10}(x) ). ( \log(x) ) grows slower than ( \ln(x) )? Wait, no: ( \ln(x) ) and ( \log(x) ) (base 10) – actually, ( \log(x)=\frac{\ln(x)}{\ln(10)} ), so ( \ln(x) ) grows faster than ( \log(x) )? Wait, no, the coefficient ( \frac{1}{\ln(10)} \approx 0.434 ) is less than 1, so ( \log(x) ) is a vertical compression of ( \ln(x) ), meaning ( \ln(x) ) is steeper. Wait, the red (R) and black (K) curves: red is steeper? Wait, no, in the graph, red (R) and black (K) – let's check domains. ( \ln(x) ) and ( \log(x) ) have domain ( x>0 ). The red curve (R) and black (K): ( \ln(x) ) has a vertical asymptote at ( x=0 ), and ( \log(x) ) too. Wait, actually, ( \ln(x) ) and ( \log(x) ): let's check the rate. ( \ln(x) ) and ( \log(x) ) – ( \ln(x) ) is ( \log_e(x) ), ( \log(x) ) is ( \log_{10}(x) ). Since ( e \approx 2.718 ), ( \ln(x) ) grows faster than ( \log(x) ) (because base ( e ) is smaller than base 10? Wait no, smaller base (closer to 1) makes log grow faster? Wait, no: for ( a < b ), ( \log_a(x) > \log_b(x) ) for ( x > 1 ), and ( \log_a(x) ) grows faster. Wait, ( \ln(x)=\log_e(x) ), ( e \approx 2.718 ), ( 10 > e ), so ( \log_e(x) ) (ln x) grows faster than ( \log_{10}(x) ) (log x). So the steeper logarithmic curve is ( \ln(x) ), which should match red (R)? Wait no, in the graph, red (R) and black (K): red is steeper? Wait, the red curve (R) and black (K) – let's see the points. Wait, maybe I mixed up. Wait, the exponential curves: green (G) is ( 10^x ), blue (B) is ( e^x ). Logarithmic: ( \ln(x) ) (base e) and ( \log(x) ) (base 10). Since ( \ln(x) ) has a larger derivative at ( x=1 ) ( ( \frac{d}{dx}\ln(x)|{x=1} = 1 ), ( \frac{d}{dx}\log(x)|{x=1} = \frac{1}{\ln(10)} \approx 0.434 ) ), so ( \ln(x) ) is steeper, so ( \ln(x) ) matches red (R), and ( \log(x) ) matches black (K). Wait, but in the original selection, let's re - check:

• ( e^x ): blue (B) (since ( 10^x ) is steeper, green (G) is ( 10^x ))
• ( 10^x ): green (G)
• ( \ln(x) ): red (R) (steeper log)
• ( \log(x) ): black (K) (less steep log)

So the correct matches are:

• ( e^x ): b. blue (B)
• ( 10^x ): a. green (G)
• ( \ln(x) ): c. red (R)
• ( \log(x) ): d. black (K)

Answer:

• ( e^x ): b. blue (B)
• ( 10^x ): a. green (G)
• ( \ln(x) ): c. red (R)
• ( \log(x) ): d. black (K)