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.

I agree with Peter Taylor.
You get:
\begin{equation}
\frac{3x}{(2x+1)(11-4x)}\ln(-2x) = \frac{1}{3x-1}\ln(7x+2)
\end{equation}
Graphing gives about -0.15649: http://bit.ly/q5Rn4l
I don't think you can solve it exactly. A rational solution is possible but unlikely in general. Maybe you're meant to just try a whole bunch of rational numbers...
