finding value of x for an equation

if we have an equation of form $y=x^{nx+1}$ and if we are given the values of $y$ and $n$ then how can one find $x$? I have reduced the equation to $\log(y)/\log(x)=nx+1$ but can't proceed further. Is there some kind of standard equation? Thanks

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If you know about differentiation then you should look up "Newton's Method." $x=2/n$ can't possibly be correct, even as an approximate answer; for one thing, it doesn't involve $y$. –  Gerry Myerson Jun 16 '12 at 22:15
A quick search on Google reveals the page http://mathforum.org/library/drmath/view/70483.html, where Doctor Vogler points out that the function $f(X) = x^x$ is not injective, since $$(\frac{1}{2})^{(1/2)} = (\frac{1}{4})^{(1/4)}.$$ However, he points out that it is possible to restrict the domain of the function so that it is injective. Nevertheless, I'm inclined to think that due to this observation, no notation may have been invented specifically for the inverse of this function, unlike other functions such as $f(x) = e^x$ which have inverses like $f^{-1}(x)=ln(x)$.