Question

match each equation with a graph above:

• $ln(x)$ a. black (k)
• $e^{x}$ b. blue (b)
• $log(x)$ c. red (r)
• $10^{x}$ d. green (g)

Explanation:

Step1: Analyze the domain and behavior of \(y = \ln(x)\)

The function \(y=\ln(x)\) has domain \(x>0\). As \(x\to0^{+}\), \(y\to-\infty\), and as \(x\to+\infty\), \(y\to+\infty\). The slope of \(y = \ln(x)\) is given by \(y'=\frac{1}{x}\), which is positive for \(x > 0\) and decreases as \(x\) increases. The graph of \(y=\ln(x)\) is similar to the blue - colored graph (B).

Step2: Analyze the domain and behavior of \(y = e^{x}\)

The function \(y = e^{x}\) has domain \((-\infty,\infty)\). When \(x = 0\), \(y=e^{0}=1\). As \(x\to-\infty\), \(y\to0\), and as \(x\to+\infty\), \(y\to+\infty\). The slope \(y'=e^{x}>0\) for all \(x\), and the function is concave up (\(y''=e^{x}>0\)). The graph of \(y = e^{x}\) is similar to the black - colored graph (K).

Step3: Analyze the domain and behavior of \(y=\log(x)\)

The function \(y = \log(x)\) (assuming base \(10\)) has domain \(x>0\). As \(x\to0^{+}\), \(y\to-\infty\), and as \(x\to+\infty\), \(y\to+\infty\). The slope \(y'=\frac{1}{x\ln(10)}\), which is positive for \(x > 0\) and decreases as \(x\) increases. Since \(\ln(10)>1\), \(y = \log(x)\) grows more slowly than \(y=\ln(x)\). The graph of \(y=\log(x)\) is similar to the green - colored graph (G).

Step4: Analyze the domain and behavior of \(y = 10^{x}\)

The function \(y = 10^{x}\) has domain \((-\infty,\infty)\). When \(x = 0\), \(y = 10^{0}=1\). As \(x\to-\infty\), \(y\to0\), and as \(x\to+\infty\), \(y\to+\infty\). The slope \(y'=10^{x}\ln(10)>0\) for all \(x\), and the function is concave up (\(y''=10^{x}(\ln(10))^{2}>0\)). Since \(\ln(10)>1\), \(y = 10^{x}\) grows faster than \(y=e^{x}\). The graph of \(y = 10^{x}\) is similar to the red - colored graph (R).

Answer:

\(\ln(x)\)-b. blue (B); \(e^{x}\)-a. black (K); \(\log(x)\)-d. green (G); \(10^{x}\)-c. red (R)