Consider the sequence of functions $f_n:[a,\infty)\to \mathbb{R}, f_n=x-a-\ln(\frac{nx+1}{na+1})$. For strictly positive $a$, analyze the uniform convergence of the sequence.
It's clear that $-\ln(\frac{1+nx}{1+na})\to -\ln(x/a)$ as $n\to\infty$
Is this enough to show the convergence of the sequence?
Let $f : x \in [a,+\infty) \mapsto x - a -\displaystyle\ln\left(\frac{x}{a}\right)$.
$(f_n)_n$ converges pointwise to $f$ on $[a,+\infty)$, let's show $(f_n)_n$ converges uniformly to $f$ on $[a,+\infty)$ too.
Let $x \in [a,+\infty)$. We then have, for $n \geq 1$ (it's fine to leave out $n = 0$ since what interests us is the behaviour when $n$ tends to $+\infty$) : $$\begin{split}\left|f_n(x) - f(x)\right| &= \left|-\ln\left(\frac{nx+1}{na+1}\right) + \ln\left(\frac{x}{a}\right)\right|\\ &= \left|-\ln\left(\frac{nx+1}{na+1}\right) + \ln\left(\frac{nx}{na}\right)\right|\\ &= \left|\ln\left(\frac{nx}{nx+1}\right) - \ln\left(\frac{na}{na+1}\right)\right|\\ &= \left|\ln\left(1 - \frac{1}{nx+1}\right) - \ln\left(1 - \frac{1}{na+1}\right)\right|\\ &= \ln\left(1 - \frac{1}{nx+1}\right) - \ln\left(1 - \frac{1}{na+1}\right)\end{split}$$ This expression is increasing in $x$ by usual operations on monotonic functions, and: $$\ln\left(1 - \frac{1}{nx+1}\right) - \ln\left(1 - \frac{1}{na+1}\right) \xrightarrow[x \to +\infty]{} \ln(1) - \ln\left(1 - \frac{1}{na+1}\right) = - \ln\left(1 - \frac{1}{na+1}\right)$$ Therefore: $$0 \leq |f_n(x) - f(x)| \leq - \ln\left(1 - \frac{1}{na+1}\right)$$ Finally, since the right side is constant with regards to $x$ and converges to $-\ln(1) = 0$ when $n$ tends to $+\infty$, we obtain, by the sandwich theorem: $$\begin{align*}\|f_n - f\|_{\infty,[a,+\infty)} \xrightarrow[n \to +\infty]{} 0\\ {}\end{align*}$$ Thus, $(f_n)_n$ converges uniformly to $f$ on $[a,+\infty)$.