# Solving $n^a = x(n^x)$ for $x$

Okay. Let me just get straight to the point. I have a formula, $n^a = x(n^x)$. What I'm trying to do is make $x$ the subject of the formula. In other words, I want $x$ to express in terms of $a$ and $n$ only. It occured to me that this problem seemed rather impossible, but I'm no expert, you all are. So, here's my question. Is it possible, using any current mathematical hocus pocus, to express $x$ in terms of $a$ and $n$ only? If so, which area of maths do we have to get ourselves into? Can calculus help?

Here are some facts and informations that I've gathered from this formula and feel free to correct me if I'm wrong:

1) $a \geq x$ (obviously)

2) $a = x = 1$ (even more obvious)

3) $n$ is a constant, $a$ is the independent variable and $x$ is the dependent variable.

4) All value of $x$ are real numbers

5) When $a = 0$, $n$th root of $n = 1/x$

That's all the info I currently have. I really wish my question receives some decent answers. Thank you all for helping a poor little boy. I'm 15 in case you're wondering. I just realized I couldn't attach a picture because I don't have enough reputation :(

• If it is really obvious that $a = x = 1$, then simply write $x = 1$. Are you absolutely sure that this is obvious? What you are really trying to ask is how to solve the equation $$n^a = x \cdot x^n$$ for $x$. – Hans Hüttel Nov 29 '16 at 20:43
• @HansHüttel That is trivial, since the RHS simplifies nicely, unlike the above problem. – Simply Beautiful Art Nov 29 '16 at 20:43
• Yes, but that is the implication that if $a=x$, then $a = 1$. – Hans Hüttel Nov 29 '16 at 20:45

One requires the use of the Lambert W function, which is required in step 3. The solution is given as follows,

$$n^a=xn^x=xe^{x\ln(n)}\tag1$$

$$n^a\ln(n)=x\ln(n)e^{x\ln(n)}\tag2$$

$$W\left(n^a\ln(n)\right)=x\ln(n)\tag3$$

$$x=\frac{W\left(n^a\ln(n)\right)}{\ln(n)}\tag4$$

• So basically to OP: it can't be written in terms of normal functions, as in with functions that are normally taught in high school courses. – Skeleton Bow Nov 29 '16 at 20:43