Find x to keep the equality $\sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}$ $$\mbox{ Find }x \in \mathbb{Q} \mbox{ to keep the equality: } \sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}$$
I tried to write the roots using powers:
\begin{align*}\sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}&\Rightarrow
[(-2x)^{\frac{3x}{11-4x}}]^{\frac{1}{2x+1}}=(7x+2)^{\frac{1}{3x-1}}\\ 
&\Rightarrow (-2x)^{\frac{\frac{3x}{11-4x}}{2x+1}}=(7x+2)^{\frac{1}{3x-1}}\\ 
&\Rightarrow (-2x)^{\frac{3x}{(11-4x)(2x+1)}}=(7x+2)^{\frac{1}{3x-1}}\\ 
&\Rightarrow (-2x)^{\frac{3x}{-8x^2+18x+11}}=(7x+2)^{\frac{1}{3x-1}}
\end{align*}
I hope I did it right until this point. But I've stuck here. Can someone help me?
Thanks.

Take logs and then use your favourite numerical root-finding algorithm. Hint: the logarithms are only both real for $-\frac{2}{7} < x < 0$.
