The vertical asymptote of the function y=log(x-2)+10 is:?
A.) x=2
B.) x=10
C.)x=-2
D.) x=-10

Question

The vertical asymptote of the function y=log(x-2)+10 is:?
A.) x=2
B.) x=10
C.)x=-2
D.) x=-10

anonymous · Answer

Try plugging in the given $x$ values and see what happens.

samsan9 · Answer

2would make the inside 0 and 10 would make 8 -2 would make it -4 and -10 would make it -12

anonymous · Answer

Right, but consider the domain of the logarithm. For any base $b$, $\log_b f(x)$ is only defined when $f(x)>0$. This means that whatever the "inside" of the logarithm is, it must be greater than zero. This also means asymptotes occurs for whichever values of $x$ make the inside function 0.

Which value of $x$ would it be for $\log(x-2)$?

samsan9 · Answer

-2 since it would make it less than 0?

anonymous · Answer

-2 would make it less than 0, but an asymptote occurs for the $x$ that makes it exactly 0.

For example, $\ln x$ is undefined at $x=0$ because the inside function, $x$ is 0 when $x=0$.

samsan9 · Answer

oh wait my bad it would 10

anonymous · Answer

Not 10 either, since
$$\log(10-2)+10=\log8-2$$
is defined.

What makes $x-2$ equal to 0?

samsan9 · Answer

so just 2 ?

anonymous · Answer

Yes, when $x=2$ you have $\log(2-2)+10=\log0+10$, but $\log0$ is undefined.