You should try to look for an equivalent of $f(a)$ or $\ln(f(a))$:
First as you said f is an increasing function, and: $\lim f = + \infty$
So you have: $f(a)[1+\ln(f(a))]=a \implies f(a)\ln(f(a))$~$a$
Here it means: $f(a)\ln(f(a))= a + o(a) = a[1+ o(1)]$
You might want to consider the $\ln$ on both sides
Edit
I'll detail o() and ~ notations so that you can understand here why it can be nice to use it:
Let f and g be two real valued functions:
f(x) = o(g(x)) when $x \rightarrow + \infty $ means: $\forall \epsilon >0 , \exists A \in R :x>A \implies |f(x)|< \epsilon|g(x)|$
If g never cancels, it's equivalent to: $\lim \frac{f(x)}{g(x)} = 0 , x \rightarrow + \infty$
Likewise you can define this notion when $x \rightarrow a , a \in R$ if for instance g diverges in a. f is said to be negligible compared to g.
- f(x) ~ g(x) when $x \rightarrow + \infty$ means : $f(x) = g(x) + o(g(x))$
When g never cancels it's the same as: $\lim \frac{f(x)}{g(x)} = 1 , x \rightarrow + \infty$
f is then said to be equivalent to g in $+ \infty$.
Likewise, the notion extends to the case where $x \rightarrow a , a \in R$.
Now since you're not familiar with this i can show you why it's nice to use it sometimes, like here:
$f(a)[1+\ln(f(a))]=a , f(a) \rightarrow +\infty$ when $a \rightarrow +\infty$:
1 is then negligible compared to ln(f(a)), and:
$1= \frac{f(a)[1+\ln(f(a))]}{a} = \lim_{a\rightarrow +\infty} \frac{f(a)\ln(f(a))}{a}$
So you have: $f(a)\ln(f(a)) \sim_{a\rightarrow +\infty} a$
So using the above definition you end up with what i had :
$f(a)\ln(f(a))= a + o(a) = a[1+ o(1)] $
where $o(1)$ is a function that verifies: $o(1) \rightarrow 0$ when $a\rightarrow +\infty$
See the nice thing is that you can manipulate an equation easily now, so if you use $\ln$ on both sides:
$\ln[f(a)\ln(f(a))] = \ln(f(a))+ \ln(\ln(f(a))) = \ln(a) + \ln[1+o(1)]$
You know that $\ln$ is continuous, and $\ln(1)=0$ so $\ln[1+o(1)]\rightarrow_{a\rightarrow +\infty} \ln(1)=0 \implies \ln[1+o(1)]=o(1)$ , when $a\rightarrow +\infty$
So you get: $\ln(f(a))+ \ln(\ln(f(a))) = \ln(a) + \ln[1+o(1)] = \ln(a) + o(1)$
Finally, since: $\frac{\ln x}{x}\rightarrow 0$ , when $x\rightarrow +\infty$ :
$\frac{\ln(\ln(f(a)))}{\ln f(a)} \rightarrow 0$ when $a\rightarrow +\infty$
So the equality writes:
$\ln(f(a))+ \ln(\ln(f(a))) =\ln(f(a))+o(\ln(f(a))) = \ln(a)+o(1) \implies \ln(a) = \ln(f(a)) + o(\ln(f(a))) -o(1) =\ln(f(a)) +o(\ln(f(a))) $
Since o(1) is also negligible compared to $\ln(f(a))$ so it is a $o(\ln(f(a)))$
Hence you have: $\ln(a)=\ln(f(a)) +o(\ln(f(a)))$ and this means exactly here that:
$$\lim_{a\rightarrow +\infty} \frac{\ln(a)}{\ln(f(a))}=1$$
Now you can find the limit...
It's a bit long sorry but hopefully you will see why this can be powerful once you're comfortable with the notions :)