Let $0<\delta<1$. For any $a>0$, there exists a unique $x_a>0$ that solves \begin{align} x(1+\delta)^{2x} = a. \end{align} Now consider the case when $a$ becomes large, i.e., $a\to\infty$. It follows that $x_a\to\infty$, but is there a way to determine the order of $x_a$? That is, how fast $x_a$ grows as $a\to\infty$?
P.S. If it helps, we can assume that $\delta$ is small, i.e., much closer to 0 than 1.
Inverting to solve for $x$ we have that
$$2\log(1+\delta) x e^{2\log(1+\delta)x} = 2\log(1+\delta)a \implies x = \frac{W(2\log(1+\delta)a)}{2\log(1+\delta)}$$
For small $\delta$ we can safely conclude
$$x \sim \frac{W(2\delta a)}{2\delta}$$
and from here since $\delta$ is fixed and we are taking $a$ arbitrarily large we can use the known asymptotic for product log
$$W(x) \sim \log x - \log \log x$$
to get
$$x \sim \frac{\log(2\delta a) - \log \log(2\delta a)}{2\delta}$$
Note that using the asymptotic $(1+\delta)^{2x}\sim 1 + 2\delta x$ at the beginning is not valid.