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.
$\require{begingroup} \begingroup$ $\def\W{\operatorname{W}}\def\e{\mathrm{e}}$
\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.
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