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

May 21, 2018

Domain: $\left(1 , \infty\right)$

$x$-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 $f \left(x\right)$ and $f \left(x + k\right)$ 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 $\ln \left(x - 1\right)$ is identical to the graph of $\ln \left(x\right)$, except it is translated one unit right.

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

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

Finally, since $\ln \left(x\right)$ has a vertical asymptote at $x = 0$, $\ln \left(x - 1\right)$ 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]}