= Solution
Write $\theta(x)=\sum_{p\le x}\log p$ for the <Chebyshev theta function> and $\psi(x)=\sum_{n\le x}\Lambda(n)$ for the <Second Chebyshev function>, where $\Lambda$ is the <Von Mangoldt function>. We first establish the <Chebyshev estimate>. Every <prime> $n<p\le2n$ divides the <central binomial coefficient> $\binom{2n}{n}$, so
$$
\theta(2n)-\theta(n)\le\log\binom{2n}{n}\le2n\log2.
$$
Apply this at successive powers of two and use <monotonicity> to obtain $\theta(x)=O(x)$. Since
$$
\psi(x)=\sum_{r\ge1}\theta(x^{1/r}),
$$
the terms $r\ge2$ contribute $O(\sqrt x\log x)$ and therefore $\psi(x)=O(x)$. This <Chebyshev estimate from central binomial coefficients> is all the information about the distribution of <primes> needed here.
The <Von Mangoldt divisor identity> $\log n=\sum_{d\mid n}\Lambda(d)$ gives the <factorial proof of Mertens first theorem>. With $N=\lfloor x\rfloor$, interchange the finite sums to obtain
$$
\log N!=\sum_{d\le x}\Lambda(d)\left\lfloor\frac xd\right\rfloor
=x\sum_{d\le x}\frac{\Lambda(d)}d+O(\psi(x)).
$$
Comparison of $\sum_{n\le N}\log n$ with $\int_1^N\log t\,dt$ gives $\log N!=N\log N-N+O(\log N)$. Dividing the preceding identity by $x$, using $N=x+O(1)$ and the <Chebyshev estimate>, gives
$$
\sum_{d\le x}\frac{\Lambda(d)}d=\log x+O(1).
$$
The contribution of higher <prime powers> is bounded independently of $x$, because
$$
\sum_p\sum_{r\ge2}\frac{\log p}{p^r}
=\sum_p\frac{\log p}{p(p-1)}
\le\sum_{n\ge2}\frac{\log n}{n(n-1)}<\infty.
$$
Thus the <Mertens first theorem> is
$$
\boxed{A(x):=\sum_{p\le x}\frac{\log p}{p}=\log x+O(1).}
$$
For the <square-root logarithmic sum over primes>, apply <partial summation> to $A(t)$ with weight $(\log t)^{-1/2}$. The lower endpoint is understood as $A(2^-)=0$, so
$$
\sum_{p\le x}\frac{\sqrt{\log p}}p
=\frac{A(x)}{\sqrt{\log x}}
+\frac12\int_2^x\frac{A(t)}{t(\log t)^{3/2}}\,dt.
$$
The main terms are $\sqrt{\log x}$ and $\sqrt{\log x}-\sqrt{\log2}$. The bounded error in $A(t)$ produces a bounded integral, since $\int_2^\infty dt/(t(\log t)^{3/2})<\infty$. Therefore
$$
\boxed{\sum_{p\le x}\frac{\sqrt{\log p}}p=2\sqrt{\log x}+O(1).}
$$
Neither argument uses the <Prime number theorem>.
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