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I don't know much about logarithms and Lambert $W$ function, but I managed to solve simplified versions, like $x^x = c$ and $(x^b)^x = c$.

However, initial equation turns into $a^x(e^{ln(x)}ln(x))^b = c$, where I can't get rid of $a^x$ to get the solution.

Wolfram Alpha gives me the solution, I'm a little concerned about: $x = \frac{a\log(2)}{3\cdot W(32/3 \cdot 2^{2/3} \cdot a\log(2))}$.

Can you explain how Wolfram Alpha got (if it's right) the solution or the basic principle for solving these?

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The basic principle is to manipulate the equation to get into the form $c=ze^z$, where $c$ is a constant, and $z$ is a function of $x$. Substituting variables can often help in doing this.

$$(ax^b)^x = c \Longrightarrow $$$$ ax^b = c^\frac{1}{x} \Longrightarrow $$$$a^\frac{1}{b}x = c^\frac{1}{bx}$$

Let $y=\frac{1}{x}$ then $$a^\frac{1}{b}y^{-1} = c^\frac{y}{b} \Longrightarrow $$$$ a^\frac{1}{b}y^{-1} = e^\frac{y\ln{c}}{b} \Longrightarrow $$$$ a^\frac{1}{b} = ye^\frac{y\ln{c}}{b} \Longrightarrow $$$$ \frac{a^\frac{1}{b}\ln{c}}{b} = \frac{\ln{c}\;y}{b}e^\frac{y\ln{c}}{b} \Longrightarrow $$$$ W\left(\frac{a^\frac{1}{b}\ln{c}}{b}\right) = \frac{\ln{c}\;y}{b} \Longrightarrow $$$$ y = \frac{b}{\ln{c}}W\left(\frac{a^\frac{1}{b}\ln{c}}{b}\right) \Longrightarrow $$$$ x = \frac{\ln{c}}{bW\left(\frac{a^\frac{1}{b}\ln{c}}{b}\right)} $$

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  • $\begingroup$ Wow, this is so simple! Thank you. $\endgroup$ – Supreme Entropy May 19 '17 at 16:08
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Basically, the function $W(x)$ is the inverse of $f(x)=e^xx$, so you have the identity $W(e^xx)=x$. For example, if $x^x=c$, then $$ W(\ln c)=W(\ln x^x)=W(x\ln x)=W(e^{\ln x}\ln x)=\ln x $$ Hence $x=e^{W(\ln c)}$.

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  • $\begingroup$ $ln(c) = ln( (ax^b)^x) = x (ln(a) +x ln(b))$ $\endgroup$ – Χpẘ May 4 '17 at 2:30

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