= Solution
For every <prime number> $p$ with $n<p\le2n$, the numerator $(2n)!$ contains one factor $p$, whereas neither copy of $n!$ contains one. Consequently the product of these <prime numbers> divides the <binomial coefficient> $\binom{2n}{n}$. Since this <binomial coefficient> is one positive summand of $\sum_{j=0}^{2n}\binom{2n}{j}=2^{2n}$, and other summands are positive,
$$
\theta(2n)-\theta(n)=\log\prod_{n<p\le2n}p\le\log\binom{2n}{n}<2n\log2.
$$
For $x\ge2$, choose $k$ with $2^{k-1}<x\le2^k$. The <Chebyshev function> is increasing, so telescoping gives
$$
\theta(x)\le\theta(2^k)=\sum_{j=1}^k\bigl(\theta(2^j)-\theta(2^{j-1})\bigr)<(2^{k+1}-2)\log2<4x\log2.
$$
For $x=1$ the inequality follows from $\theta(1)=0$. Thus $\boxed{\theta(x)<4(\log2)x}$ throughout the stated range.
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