Below i have an equation where $H$ and $k$ is a positive real integers and the $t$ is the changing variable.
$$t^{k-1} e^{-t} = H$$
So when I solve this on Maple in order to isolate for $t$ i get the result presented below:
$$t = -(k-1)LambertW\Bigg(-\frac{e^{-\frac{ln H}{k-1}}}{k-1}\Bigg)$$
My question is how does Maple arrive to this solution? What is confusing me is the $LambertW$ function, what is the condition such that i obtain the following solution?
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Not sure, how exactly the Maple algorithms work, but typical transformations of the given equation to the form $u\,\e^u=v$ in order to apply the Lambert $\W$ function looks something like this:
\begin{align} t^{k-1} e^{-t} &= H ,\\ t\exp\left(-\frac t{k-1}\right) &= H^{\frac 1{k-1}} \quad\text{note that k=1 is a special simpler case} ,\\ -\frac t{k-1}\,\exp\left(-\frac t{k-1}\right) &= -\frac 1{k-1}\,H^{\frac 1{k-1}} \end{align}
At this point we have the desired form $u\,\e^u=v$ of the original equation, where
\begin{align} u&=-\frac t{k-1} ,\\ v&=-\frac 1{k-1}\,H^{\frac 1{k-1}} \end{align}
and we can apply the Lambert $\W$ function to get $u$ on the left as
\begin{align} \W(u\,\e^u)&=\W(v) ,\\ u&=\W(v) , \end{align}
so
\begin{align} \W\left(-\frac t{k-1}\,\exp\left(-\frac t{k-1}\right)\right) &= \W\left(-\frac 1{k-1}\,H^{\frac 1{k-1}}\right) ,\\ -\frac t{k-1} &= \W\left(-\frac 1{k-1}\,H^{\frac 1{k-1}}\right) ,\\ t &= (1-k)\,\W\left(-\frac 1{k-1}\,H^{\frac 1{k-1}}\right) . \end{align}
At tis point we have the solution of the original equation in terms the Lambert $\W$ function and we need to make one more important step: it's time to analyze the argument $v$ of $\W$ to find out the number of the real solutions.
It is well-known that
\begin{align} v<-\frac1\e\quad&\Rightarrow\quad\text{no real solutions} ,\\ v\ge0\quad&\Rightarrow\quad\text{one real solution, }\Wp(v) \text{ or just } \W(v) ,\\ v\in(-\tfrac1\e,0) \quad&\Rightarrow\quad\text{two real solutions, }\Wp(v)\in(-1,0) \text{ and } \Wm(v)<-1 ,\\ v=-\tfrac1\e \quad&\Rightarrow\quad\text{a special case, one real solution, } \Wp(-\tfrac1\e)=\Wm(-\tfrac1\e)=-1 . \end{align}
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