A question about $|X|\le d(X)^{\chi(X)}$ It is well known that the following cardinal inequality $|X|\le d(X)^{\chi(X)}$
holds,
where $d(X)$ denotes density of $X$ and $\chi(X)$ denotes character of $X$.
Thus we have 

$|X|\le d(X)^\omega$ if $X$ is first countable.

My question is this:

In the above conclusion, could the condition "first countable" be replaced by "countable tightness and countable pseudocharacter"?, i.e., do we have 
  $|X| \le d(X)^\omega$ if $X$ has countable tightness and countable pseudocharacter?

Thanks ahead.
 A: I now do have an argument for what you originally asked:
Assume $X$ is $T_3$ and has countable tightness and pseudocharacter.
Let $D$ be a dense subset of size $d(X)$.
As each $x \in X$ is in $\overline{D}$, and $X$ has countable tightness 
we can choose for each $x \in X$ a countable subset $D_x$ of $D$ such that
$x \in \overline{D_x}$.
As $X$ has countable pseudocharacter, and is regular, we can pick for each $x$ a countable sequence $(U_n(x))$ of open sets such that $$\{x\} = \bigcap_n \overline{U_n(x)}\text{.}$$
Now define $$f: X \rightarrow ([D]^\omega)^\omega \text{ by } f(x) = (U_n(x) \cap D_x)_{n \in \omega}$$
Claim: $f$ is 1-1. For suppose $x \neq y$. Then there exists an $n$ such that $y \notin \overline{U_n(x)}$. As clearly for any $n$, $y \in \overline{U_n(y) \cap D_y}$ ,we see that for that first $n$: $$y \notin \overline{U_n(x) \cap D_x} \subseteq \overline{U_n(x)} \text{ but } y \in \overline{U_n(y) \cap D_y}$$ 
meaning that $f(x)_n \neq f(y)_n$ (the closures of these sets are different so the sets are different too) so $f(x) \neq f(y)$.
As $|([D]^\omega)^\omega| = |D^\omega| = d(X)^\omega$, we have that $|X| \le d(X)^\omega$ for those spaces.
Analysing the proof we see that in fact that for regular spaces $X$:
$$|X| \le d(X)^{t(X)\psi(X)}$$
using sequences of length $\psi(X)$ of subsets of $D$ of size $t(X)$. As $t(X)\psi(X) \le \chi(X)$, this indeed generalises you original $d(X)^{\chi(X)}$ bound for $|X|$.
(after some thought on necessity of $T_3$-ness):
Define $X = \{\mathscr{F}: \mathscr{F} \text {a free ultrafilter on }\omega\} \cup \omega$. It is well known that $|X| \ge 2^\mathfrak{c}$. Define a topology by local bases: all $n \in \omega$ are isolated, while a neighbourhood of $\mathscr{F}$ is of the form $\{\mathscr{F}\} \cup A$, where $A \in \mathscr{F}$. It's I( I think ) easy to check that $X$ is Hausdorff, $t(X) = \omega = \psi(X)  =d(X)$, but $X$ does not obey the bound. So $X$ cannot be regular, in fact. (The proven bound actually shows this, it's similar to using Jones' lemma to show non-normality of spaces).      
