for the given logarithmic function, complete parts a through c. a. determine 3 points that lie on the graph…

Explanation:

Step1: Recall the property of $y = \ln x$

For $y=\ln x$, when $y = - 1$, we solve $\ln x=-1$. By the definition of the natural - logarithm ($y = \ln x\Leftrightarrow x = e^{y}$), we have $x = e^{-1}=\frac{1}{e}$. For the function $f(x)=\ln(x - 1)$, when $x=\frac{1}{e}+1$, $f(\frac{1}{e}+1)=\ln((\frac{1}{e}+1)-1)=\ln(\frac{1}{e})=-1$. So the point on $y = \ln x$ is $(\frac{1}{e},-1)$ and the corresponding point on $f(x)=\ln(x - 1)$ is $(\frac{1}{e}+1,-1)$.

Step2: Recall the property of $y = \ln x$

For $y=\ln x$, when $y = 0$, we solve $\ln x = 0$. Since $\ln x=0\Leftrightarrow x = e^{0}=1$. For the function $f(x)=\ln(x - 1)$, when $x=1 + 1=2$, $f(2)=\ln(2 - 1)=\ln(1)=0$. So the point on $y=\ln x$ is $(1,0)$ and the corresponding point on $f(x)=\ln(x - 1)$ is $(2,0)$.

Step3: Recall the property of $y = \ln x$

For $y=\ln x$, when $y = 1$, we solve $\ln x=1$. Since $\ln x = 1\Leftrightarrow x=e^{1}=e$. For the function $f(x)=\ln(x - 1)$, when $x=e + 1$, $f(e + 1)=\ln((e + 1)-1)=\ln(e)=1$. So the point on $y=\ln x$ is $(e,1)$ and the corresponding point on $f(x)=\ln(x - 1)$ is $(e + 1,1)$.

Answer:

Points that lie on the graph of $y=\ln x$	Corresponding points that lie on the graph of $f(x)=\ln(x - 1)$
$(\frac{1}{e},-1)$	$(\frac{1}{e}+1,-1)$
$(1,0)$	$(2,0)$
$(e,1)$	$(e + 1,1)$