<s>[INST] Finding expression for variable x in an equation (use Lambert function?) Let $a$, $b$ and $c$ be constants. How can one find an expression for variable $x$ in the following equation? $$\frac{a\cdot (b+x)}{c} = (1+\frac{a\cdot x}{c}) \cdot \ln(1+\frac{a\cdot x}{c})$$
From my research, I am being suggested to use the Lambert function, but I still feel quite helpless! It seems very complex.
 [/INST] $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\e{\mathrm{e}}$
\begin{align} 
\frac{a\cdot (b+x)}{c} = (1+\frac{a\cdot x}{c}) \cdot \ln(1+\frac{a\cdot x}{c})
\tag{1}\label{1}
\end{align} 
Indeed, the solution to equation \eqref{1} 
can be expressed in terms of the Lambert W function.
The usual approach in this case is to transform
original equation to the form
\begin{align} 
t\cdot\exp(t)&=\dots
\tag{2}\label{2}
,
\end{align}
where the right-hand side of \eqref{2}
does not depend on $t$,
and then apply the Lambert W function
\begin{align} 
\W(t\cdot\exp(t))&=\W(\dots)
\tag{3}\label{3}
\\
\text{to get }\quad
t&=\W(\dots)
\tag{4}\label{4}
.
\end{align}
First, simplifying \eqref{1} using substitution
\begin{align} 
y&=1+\frac{ax}c
\tag{5}\label{5}
,\\
x&=\frac ca\cdot(y-1)
\tag{6}\label{6}
,
\end{align}
we get
\begin{align} 
\ln(y)\cdot y&=\frac{ab}c-1+y
,\\
\ln(y)\cdot y-y&=\frac{ab}c-1
,\\
(\ln(y)-1)\cdot y&=\frac{ab}c-1
,\\
\ln(\tfrac y\e)\cdot y&=\frac{ab}c-1
,\\
\ln(\tfrac y\e)\cdot \tfrac y\e&=\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)
,\\
\ln(\tfrac y\e)\cdot 
\exp\left(\ln(\tfrac y\e)\right)&=\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)
\tag{7}\label{7}
.
\end{align}
Now we have \eqref{7} in the form of \eqref{2}, hence
\begin{align} 
\W\left[\ln(\tfrac y\e)
\cdot 
\exp\left(\ln(\tfrac y\e)\right)
\right]
&=
\W\left[\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)\right]
,\\
\ln(\tfrac y\e)
&=
\W\left[\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)\right]
\tag{8}\label{8}
,
\end{align}
and from here $x$ can be trivially found.
$\endgroup$
</s>