Is $\ln(x)$ uniformly continuous?

You can prove something more general:

PROP Suppose $f:[a,\infty)\to\Bbb R$ has bounded derivative. Then $f$ is uniformly continuous on its domain.

P Pick $x,y\in[a,\infty)$ arbitrarily. By the mean value theorem, we can write $$|f(x)-f(y)|=|f'(\xi)||x-y|$$

Let $M=\sup\limits_{x\in[a,\infty)}|f'(x)|$. Then $$|f(x)-f(y)|\leq M|x-y|$$

Thus, for any $\epsilon$ we may take $\delta=\frac{\epsilon}{2M}$. Note that in your case $M=1$. I only divide by $2$ to turn $\leq$ into $<$.

ADD This means, for example, that $\log x$ (over $[a,\infty)$, $a>0$), $\sin x$, $\cos x$, $x$, and similar functions are all uniformly continuous. Note, for example, that $\sin(x^2)$ is not uniformly continuous. Note that we actually prove $f$ is $1$-Lipschitz with constant $M$, so this might be of interest.

An easier argument is to note that the derivative of $\ln x$ is bounded by 1 on the interval $[1,\infty)$. Therefore $\ln x$ is Lipschitz and in particular uniformly continuous.

Alternatively, you can prove a function is uniformly continuous based off the following idea:

$f$ is uniformly continuous if and only if for any sequence $\{a_n\},\{b_n\}$

$$ \lim\left(a_n-b_n\right)=0 \Rightarrow \lim\left(f\circ a_n-f\circ b_n\right)=0. $$

Let $\{a_n\}, \{b_n\}$ satisfy our hypothesis ($\lim\left(a_n-b_n\right)=0$), then we have $\lim a_n = \lim b_n$ and so

$$ \lim\left(f\circ a_n-f\circ b_n\right) =\lim\left(\ln(a_n)-\ln(b_n)\right)=\lim\ln\left(\frac{a_n}{b_n}\right) = \ln(1) = 0. $$