I have been playing with graphs until made a nice equation
On
Graph of a concave up function intersects the graph of a concave down function on a closed interval at at most 2 points.
Let $f(x)=\frac{36}{\ln 2}\ln x$ and $g(x)=4x^4-3x^3-2x^2+x$. Then, $f''(x)=-\frac{36}{x^2\ln2}<0$ and so $f(x)$ is concave down on its domain which is $x>0$.
On the other hand, $g''(x)=48x^2-18x-4.$ The roots of $g''(x)$ are $\frac{9\pm{\sqrt{273}}}{48}.$ So, g(x) is concave up for $x\geq\frac{9+\sqrt{273}}{48}\approx0.53172$ and we conclude that there are two solutions of $f(x)=g(x)$ on this interval.
On $(0,\frac{9+\sqrt{273}}{48})$ both are concave down. We need another analysis here. Or by a careful curve sketching we can see that the graph of the quartic stays above of the logarithm. No solition here.
You can at least have a very good approximation of it.
Multiplied by $x$, the first derivative of $\text{(rhs - lhs)}$ is $$16 x^4-9 x^3-4 x^2+x-\frac{34}{\log(2)}$$ which can be solved using radicals. Converted to decimals, its positive zero is $$x_*=1.539220361840668443650\cdots$$ The second derivative test shows that it corresponds to a maximum.
Since we already needed to solve a quartic equation, expanding $$\text{(rhs - lhs)}=f(x)=x(x-1)(4x^2+x-1)-\frac{34}{\log (2)}\, \log(x)$$ around $x_*$ to fourth order, the two real approximate zero's are $$x_1=0.998345 \qquad \text{and} \qquad x_2=1.999688$$