Let $x = \frac{\ln T}{\ln 2} = 0.879146\dots$ where $T$ is the tribonacci constant, then x solves the transcendental equation,
$$4^x(2^x-1)=(2^x+1)$$
Let $x = \frac{\ln y}{\ln 2} = 1.523627\dots$ where $y$ is the real root of the cubic $y^3-y^2-4y-4=0$ (hence $x=1.5236\dots$ is the boundary of the dragon curve), then x solves,
$$4^x(2^x-1)=4(2^x+1)$$
In general, given the transcendental equation with unknown $x$ and algebraic constants $a_i,b_i,c_i,d_i$,
$$ (a_1p_1^x + b_1) (a_2 p_2^x + b_2) \dots = (c_1q_1^x + d_1) (c_2q_2^x + d_2)\dots\tag{1}$$
for the special case when $p = p_i = q_i$, then the solution to $(1)$ for $p \ne 0, 1$ is $x = \frac{\ln z}{\ln\, p}$ where $z$ is a root of the algebraic equation,
$$(a_1z + b_1) (a_2z + b_2) \dots = (c_1z + d_1) (c_2z + d_2) \dots$$
Question: Excluding using the Lambert W function, is there any other special case of the transcendental equation $(1)$ that has a simple closed-form solution?