Show: $y(t)\to\infty$ implies $x(t)\to\infty$ and $x(t)

Suppose $y(t)=\frac{1}{a}\ln(a)+\frac{1}{a}\ln(t+C)$ where $a>0$ is some constant. Then, in particular $y(t)\to\infty$ as $t\to\infty$. I want to show that for
    $$
x'(t)=e^{-ax(t)}-e^{-a y(t)},
$$
    this implies (1) $x(t)\to\infty$ as $t\to\infty$ and (2) there exists some $s>0$ such that $x(t)<y(t)$ for all $t>s$.

My argument for (1) goes as follows.
Assume by contradiction that $\lim_{t\to\infty}x(t)=X\in\mathbb{R}$, i.e. that $x$ converges.
Then, since by assumption $\lim_{t\to\infty}e^{-ay(t)}=0$, there exists some $T>0$ such that
$$
x'(t)>0~\forall t>T.
$$
But this contradicts the convergence of $x$. Hence, $x(t)\to\infty$ as $t\to\infty$.
For (2) my idea is again to use contradiction:
Suppose $x(t)\geqslant y(t)$, then $x'(t)\leqslant 0$ since $e^{-a x(t)}\leqslant e^{-a y(t)}$. Dont know if this can give a contradiction.
Or maybe since $x(t)<y(t)$ is equivalent to $x'(t)>0$ it is better to show directly that $x'(t)>0$.
 A: About your argument for (1):
Firstly, I don't think that the implication that there is some $T$ such that $x'(t)>0$ for $t>T$ is true. It could still be that $x(t)-y(t)$ alternates signs.
Moreover, even if you manage to show that $x'(t)>0$ eventually, a function being strictly increasing does not guarantee that it does not converge. Think about $f(x)=-\frac{1}{x}$ for $x>0$.
A: As to argument 1), if $x(t)\le M$, then $e^{-ax(t)}\ge e^{-aM}$ and for $t$ large enough the $e^{-ay(t)}$ term will be smaller than half of that, so that you get a positive minimal slope for $x$, contradicting the boundedness.
In general if $x'=f(y)-f(x)$ with an increasing $f$, then $y(t)$ is a transient curve and at any point in time $x$ will move towards it. This means that $x$ lags behind, and for an increasing $y$ this means that eventually the $x$-curve comes to move below the $y$-curve.

But now with the given form of $y$ you can even compute more. Set $u(t)=e^{ax(t)}$ and insert $y$ to get the linear DE
$$
u'(t)=ae^{ax(t)}x'(t)=a-\frac{u(t)}{t+C}
\\
(t+C)u(t)=\frac{a}2(t+C)^2+D\implies u(t)=\frac{a}2(t+C)+\frac{D}{t+C}
$$
which clearly is unbounded and eventually smaller than $e^{ay(t)}=a(t+C)$.
