How to solve exponential function of form $a b^x + x = c$? For instance, how would I solve:
  $3^x + x = 85$ ? 
 A: You can solve this using the product log function, its a special function, so if your not used to using them, you might just want a numeric answer. But your special case has the simple solution 4. Here is a link, http://mathworld.wolfram.com/LambertW-Function.html.
A: If we consider $x\ge 0,3^x\le 85,x\le 4$
If $x<4,3^x<81\implies 3^x+x<85$
So $x$ can be $4$ which actually satisfies the given equation.
If $x<0,3^x<1\implies -3^x>-1\implies x=85+(-3^x)>84$ which is impossible as $x<0$
A: $$a b^x + x = c$$
$$a b^x = c-x$$
$$ b^x = \frac{c-x}{a}$$
$$ e^{x\ln{b}} = \frac{c-x}{a}$$
$$ 1= \frac{c-x}{a} e^{-x\ln{b}}$$
$$  e^{c\ln{b}}= \frac{c-x}{a} e^{-x\ln{b}} e^{c\ln{b}}$$
$$  ae^{c\ln{b}}= (c-x) e^{(c-x)\ln{b}}$$
 $$  \ln{b}.a.e^{c\ln{b}}= \ln{b}(c-x) e^{(c-x)\ln{b}}$$
 $$  b^c\ln{b}.a= \ln{b}(c-x) e^{(c-x)\ln{b}}$$
$u=\ln{b}(c-x)$
$$ ue^u= b^c\ln{b}.a$$
$$ u= W(b^c\ln{b}.a)$$
where $W(x)$ is Lambert W function
$$u=\ln{b}(c-x)=W(b^c\ln{b}.a)$$
$$x=c-\frac{W(a b^c\ln{b})}{\ln{b}}$$
It is general solution of  $a b^x + x = c$
For your example:
$3^x + x = 85$
$$x=85-\frac{W(3^{85}\ln{85})}{\ln{85}}$$
