The first answer (form Shaktal)is in my opinion perfect from the mathematical point of view (solving the differential equation) as well as from the physical point of view (having all units in order).
However your probably conceive your approach more intuitive because you know this from calculating things with Interest i.e.
$N_{Money}=N_{0,Money}\times (1\pm Interest)^{t(in\;years)}$
Aside from the units being messed up, physicists like functions like $e^{something}$ because you can easily calculate with them. I.e. try to take
$\frac{d}{dt}N_{wrong}$.
$\frac{d}{dt}N_{wrong}=\frac{d}{dt}N_0\times(1-k)^t=\frac{d}{dt}N_0 e^{ln(1-k)t}$
gives you just
$\frac{d}{dt}N_0 e^{-\lambda t}$
so why the detour?
The derivate is called the activity and is very important so you have to calculate it quite often. You may also want to integrate $N_{wrong}$ sometimes which would also be really annoying.
Further considering the answer from Samuel Hapak:
The half-life (not the great game) as the "real" half-life of an element, as in
$N_{1/2}=N_0\times\left(\frac{1}{2}\right)^{t_{1/2}/\lambda}\;\;t_{1/2}\text{: half-life, }\lambda:\frac{1}{\text{(half-life)}} $
gives you a quick overview over the decay rate with an SI-unit. So it can easily be given an order of Magnitude one can relate to. So it is indeed as intuitive as it can get in my opinion. One can of course easily switch to the mathematical expression resulting from solving the differential equation shown in Shaktal's answer:
$N_t=N_0 e^{-\tilde\lambda t}$
by setting $\tilde\lambda = ln(2)\lambda $
The first of your equations is correct. You can see this in two ways. First, just look at the dimensions. In general, the argument of a logarithm should be dimensionless; only your first option is. Second, and maybe more convincingly, look at what you get when you take $\Lambda \to0$. You should be able to reproduce the standard decay equation:
\begin{equation}
N_X(t) = N_0\, e^{-\lambda\, t}~.
\end{equation}
In your first equation, the factors of $\lambda$ on the left-hand side cancel, and you get this result. With your second equation, you would get $N_X(t) = \frac{N_0}{\lambda}\, e^{-\lambda\, t}$. So that must be wrong.
As for what $t_{1/2}$ is, surely it must just be the half-life of $\mathbb{X}$ (with no creation). In particular, if $\Lambda$ is large enough, $N_X$ will actually grow, so there is no time at which half of the material is left. Since $\mathbb{Y}$ is stable, you can assume there's no relevant half-life there.
Also, your expression for $N_Y$ is correct. It's a slightly harder integration, but not too bad.
Best Answer
The chance for a fixed nucleus to decay doesn't depend on the number of nuclei. In a fixed amount of time all the nuclei have a certain chance to decay. Increasing the number of nuclei will increase the number of nuclei that decay, but that's really just what you'd expect.
It's like rolling lots dice, the number of dice showing a certain digit will be proportional to the number of dice you roll. Continuing this analogy, say you remove a die if it shows 1. This correspond to a nucleus decaying. Then at first many dice will be removed, but as the number of dice grow smaller so will the number that are removed.