7. Given the function $f(x)=x^{2}+2 x+a \ln x$, for any $t \geqslant 1$ it always holds that $f(2 t-1) \geqslant 2 f(t)-3$. Then the range of the real number $a$ is . $\qquad$
7. $a \leqslant 2$.

According to the problem, the condition of the original problem is equivalent to: for any $t \geqslant 1$, it always holds that
$$
\begin{array}{l}
2(t-1)^{2}+a \ln (2 t-1)-2 a \ln t \geqslant 0 . \\
\text { Let } u(t)=2(t-1)^{2}+a \ln (2 t-1)-2 a \ln t .
\end{array}
$$

Then $u^{\prime}(t)=4(t-1)+\frac{2 a}{2 t-1}-\frac{2 a}{t}$
$$
=\frac{2(t-1)(2 t(2 t-1)-a)}{t(2 t-1)} \text {. }
$$

Since $2 t(2 t-1) \geqslant 2$, therefore
when $a \leqslant 2$, $u^{\prime}(t) \geqslant 0$, then $u(t) \geqslant u(1)=$ 0, which holds;

when $a>2$, $u(t)$ is decreasing on the interval $\left(1, t_{0}\right)$ and increasing on the interval $\left(t_{0},+\infty\right)$, where $t_{0}=\frac{1+\sqrt{1+4 a}}{4}$ $>1$, at this time, for any $t \in\left(1, t_{0}\right)$, we have $u(t)$ $<u(1)=0$, which does not meet the requirement.
Thus, the range of the real number $a$ is $a \leqslant 2$.