Question

match each equation with a graph above

y = \ln(2 - x) \t\t a. green
y = 1.6\ln(x) \t\t\t b. blue
y = \log(1 - x) \t\t c. black
y = 1.2\log(x) \t\t d. red

Explanation:

Step1: Analyze domain of each function

1. $y=\ln(2-x)$: Domain $2-x>0 \implies x<2$, defined for all $x<2$, passes through $(1, \ln(1))=(1,0)$
2. $y=1.6\ln(x)$: Domain $x>0$, defined for positive $x$, passes through $(1, 1.6\ln(1))=(1,0)$
3. $y=\log(1-x)$: Domain $1-x>0 \implies x<1$, defined for $x<1$, passes through $(0, \log(1))=(0,0)$
4. $y=1.2\log(x)$: Domain $x>0$, defined for positive $x$, passes through $(1, 1.2\log(1))=(1,0)$

Step2: Identify decreasing/increasing behavior

1. $y=\ln(2-x)$: Derivative $\frac{-1}{2-x}<0$, decreasing, $x<2$
2. $y=1.6\ln(x)$: Derivative $\frac{1.6}{x}>0$, increasing, $x>0$
3. $y=\log(1-x)$: Derivative $\frac{-1}{(1-x)\ln(10)}<0$, decreasing, $x<1$
4. $y=1.2\log(x)$: Derivative $\frac{1.2}{x\ln(10)}>0$, increasing, $x>0$

Step3: Match to graphs

• Green (increasing, $x>0$, steep): $y=1.6\ln(x)$ (matches a)
• Blue (increasing, $x>0$, gentle): $y=1.2\log(x)$ (matches b)
• Black (decreasing, $x<2$): $y=\ln(2-x)$ (matches c)
• Red (decreasing, $x<1$): $y=\log(1-x)$ (matches d)

Answer:

a. green: $y=1.6\ln(x)$
b. blue: $y=1.2\log(x)$
c. black: $y=\ln(2-x)$
d. red: $y=\log(1-x)$