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$$\mbox{ Find }x \in \mathbb{Q} \mbox{ to keep the equality: } \sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}$$

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?

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uh... $x \in \mathbb{Q}$ ? are you sure you got it right? – leonbloy Oct 4 '11 at 14:35
Yes, I'm sure it's $\mathbb{Q}$ – Daniel Oct 4 '11 at 14:36
Assume that x=a/b with a and b relatively prime, notice that most prime factors of a and b would force the two exponents to be zero or to be equal which gives a non-rational value of x, so a and b are of a very restricted form. If you want to proceed from there, you better be very sure that you have the correct problem statement. – Phira Oct 4 '11 at 16:57
Is $3x-1$ correct? I don't see how could there exist three positive integers $2x+1$, $11-4x$ and $3x-1$, with $-2/7<x<0$. (nth root ) – Américo Tavares Oct 4 '11 at 18:47
up vote 3 down vote accepted

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

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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:

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...

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