Is it possible to solve $ax + \frac{bx}{y^x} < c $ for $x$? We have the equation:
$$ax + \frac{bx}{y^x} &lt; c $$
Is it possible to rearrange it to have $x$ on its own on the RHS?
I'm not sure this can be solved unless one of a,b,c is zero.
$$ax + \frac{bx}{y^x} &lt; c, \quad  a=0\qquad
\implies \frac{bx}{y^x} &lt; c\qquad \\
\implies
x &lt; \frac{-W_n\big(-\frac{c \log y}{b}\big)}{\log y}\qquad
b,y,(c \log y)\ne0 
\quad n\in\mathbb{ Z}\tag{1}$$
$\\$
$$ax + \frac{bx}{y^x} &lt; c, \qquad b=0\quad
\implies x&lt;\frac{c}{a},\quad a\ne0\tag{2}$$
$\\$
$$ax + \frac{bx}{y^x} &lt; c, \qquad c=0\qquad
\implies a y^x + b &lt; 0\\
\implies x &lt; \frac{\log\frac{-b}{a} + 2 i \pi n}{\log y} 
\qquad   a,b,y,\log y\ne0 
\quad   n \in\mathbb{ Z} \tag{3}
$$
I'm not sure this can be solved unless one of a,b,c is zero.
$$ax + \frac{bx}{y^x} &lt; c, \quad  a=0\qquad
\implies \frac{bx}{y^x} &lt; c\qquad \\
\implies
x &lt; \frac{-W_n\big(-\frac{c \log y}{b}\big)}{\log y}\qquad
b,y,(c \log y)\ne0 
\quad n\in\mathbb{ Z}\tag{1}$$
$\\$
$$ax + \frac{bx}{y^x} &lt; c, \qquad b=0\quad
\implies x&lt;\frac{c}{a},\quad a\ne0\tag{2}$$
$\\$
$$ax + \frac{bx}{y^x} &lt; c, \qquad c=0\qquad
\implies a y^x + b &lt; 0\\
\implies x &lt; \frac{\log\frac{-b}{a} + 2 i \pi n}{\log y} 
\qquad   a,b,y,\log y\ne0 
\quad   n \in\mathbb{ Z} \tag{3}
$$