Write the statement using quantifiers I have a statement 
S: "No positive number $x$ satisfies the equation $f(x) = 5$."
Can I write the statement S as
$\nexists x \mid f(x) = 5$.
 A: Some equivalent ways of writing the statement include:
$$
\forall x \in \mathbb{R}( x\geq 0 \implies \lnot (f(x)=5))
$$
$$
\forall x \in \mathbb{R}( x\geq 0 \implies f(x)\neq 5)
$$
$$
\forall x \in \mathbb{R}_+(f(x)\neq 5)
$$
A: 
I have a statement:  $S$: "No positive number x  satisfies the equation $f(x)=5$ ."
Can I write the statement S as: $\not{∃}x\mid f(x)=5$

That says "there does not exists anything that satisfies $f(x)=5$."   Which is fine as long as the domain of discourse has already been established as "positive numbers".   Otherwise you should include the restriction.
(PS: While still viable, usage of $\;\not \exists\;$ appears to be somewhat depreciated in favour of $\;\neg\exists\;$.)
Some ways to do so are:
$$\neg \exists x: \big(x\in \Bbb R \wedge x>0 \wedge f(x)=5\big)
\\[1ex]
\neg \exists x : \big(x\in \Bbb R_+ \wedge f(x)=5\big)
\\[2ex]
\neg \exists x\in \Bbb R : \big(x>0 \wedge f(x)=5 \big)
\\[1ex]
\neg \exists x\in \Bbb R_+ : \big( f(x)=5 \big)
$$
nullUser's statements using the universal quantifier are also equivalent, via the dual negation rule: $$\neg\exists x:(x\in \mathcal A\wedge P(x)) \;\iff\; \forall x:(x\in \mathcal A \to \neg P(x))$$
A: S: "No positive number $x$ satisfies the equation $f(x) = 5$."
i.e. It is not case that there exists $ x\in \mathbb{R}^{+} $ such that $f(x) = 5$. 
$ \urcorner (\exists x\in \mathbb{R}^{+} \mid f(x) = 5)\equiv \forall  x\in \mathbb{R}^{+} ( f(x) \neq 5)$ 
Therefore for each $ x\in \mathbb{R}^{+} $, $f(x) \neq 5$. 
