Question

How do you find the domain, x intercept and vertical asymptotes of `f(x)=ln(x-1)`?

Answer 1

Domain: `(1,infty)`

`x`-intercept: `x=2`

Vertical asymptote: `x=1`

Explanation:

All of the questions can be answered easily by realising that this function is just a translated version of the original logarithm.

In fact, it is true for any function `f` that the graph of `f(x)` and `f(x+k)` only differ for a horizontal translation, `k` units to the left if `k>0`, to the right otherwise.

So, in your case, the graph of `ln(x-1)` is identical to the graph of `ln(x)`, except it is translated one unit right.

So, since the domain of `ln(x)` is `(0,\infty)`, the domain of `ln(x-1)` will be `(1,infty)`. This makes sense, since the logarithm only accepts positive inputs, and `x-1` is positive only when `x>1`.

For the same reason, since `ln(x)=0` for `x=1`, then `ln(x-1)=0` for `x=2`, which is its `x`-intercept.

Finally, since `ln(x)` has a vertical asymptote at `x=0`, `ln(x-1)` will have a vertical asymptote at `x=1`.

Here's a graph for a visualization of all the computations:
graph{ln(x-1) [-1.584, 8.416, -2.08, 2.92]}