6) One of the main characteristics of the logarithmic function is that
[tex]\text{domain}(\log _b(x))=(0,\infty)[/tex]
For b>0. Then, in our case, x-2 has to be always greater than zero; thus,
[tex]\begin{gathered} x-2>0 \\ \Rightarrow x>2 \end{gathered}[/tex]
The domain of f(x) is
[tex]\text{domain}(f(x))=x\in(2,\infty)[/tex]
Given the domain, we can calculate the range by evaluating the function in the extremes of the interval
[tex]\lim _{x\to2}f(x)=\lim _{x\to2}\log _5(x-2)-1=-\infty-1=-\infty[/tex]
Any function of the form logb(x) ->-infinite when x->0
Similarly,
[tex]\lim _{x\to\infty}f(x)=\lim _{x\to\infty}\log _5(x-2)-1=\infty-1=\infty[/tex]
Then, the range of the function is
[tex]\text{range}(f(x))=(-\infty,\infty)[/tex]
In order to calculate the range of the function, we answer the third part of the question:
As x->infinite, f(x)->infinite
And
As x->2, f(x)->-infinite
To calculate the x-intercept, set f(x)=0 and solve for x, as follows
[tex]\begin{gathered} f(x)=0 \\ \Rightarrow\log _5(x-2)-1=0 \\ \Rightarrow\log _5(x-2)=1 \\ \Leftrightarrow5^1=x-2,\text{ definition of logarithm} \\ \Rightarrow x=7 \end{gathered}[/tex]
The x-intercept is x=7
The only asymptote is when x->2 because, in that case, f(x)->-infinite.
Then, the asymptote is x=2
It's not possible to exactly draw a logarithmic function by hand; however, we can use the information about the asymptote and the x-intercept to be more precise.
The graph of the function is
Table of values of f(x) (3 different values of x)
We already found that f(7)=0, the first value of the table is then
[tex]\begin{gathered} x\to y \\ 7\to0 \end{gathered}[/tex]
(7,0)
Then,
[tex]\begin{gathered} x=3 \\ \Rightarrow f(x)=\log _5(3-2)-1=\log _5(1)-1=0-1=-1 \\ \Rightarrow(3,-1) \end{gathered}[/tex]
(3,-1)
And
[tex]\begin{gathered} x=4 \\ \Rightarrow f(x)=\log _5(4-2)-1=\log _5(2)-1=\frac{\ln(2)}{\ln(5)}-1\approx-0.56932\ldots \\ \Rightarrow(4,-0.56932\ldots) \end{gathered}[/tex]
(4, -0.56932...)
