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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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A quick search on Google reveals the page, 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 $f^{-1}(x)=ln(x)$.

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thanks Andrew for the information – pranay Jun 16 '12 at 14:15

There is no solution in terms of the standard functions of calculus (powers, roots, logs, exponentials, trig functions, inverse trig functions). There are numerical methods for getting good approximate answers (the best ones rely on Calculus). It may be possible to express solutions in terms of the Lambert-W function, which you might want to look up.

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thanks Gerry, i tried to find approximate value by differentiating both sides with respect to x . On taking double differentiation to find the value of X , I get X=2/n but i don't think its correct . – pranay Jun 16 '12 at 14:12
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

To solve an equation like this you need the generalized Lambert's function.

First of all apply natural log on both sides:

$(nx+1)\ln x=\ln y$

Now rewrite $x$ as:

$(ne^{\ln x}+1)\ln x=\ln y$

So we can rewrite the equation in the form:

$e^{\ln x}\ln x+\frac 1n \ln x=\frac 1n \ln y$

Then apply the generalized Lambert function (you can read about that here

$\ln x=W_{\frac 1n}(\frac 1n \ln y)$

$$x=e^{W_{\frac 1n}(\frac 1n \ln y)}$$

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