Solving $a=\Big(1+\frac{b}{x}\Big)^x$ for $x$ How to solve this equation for $x$?
$$a=\Bigg(1+\frac{b}{x}\Bigg)^x$$
It's not a task that I was asked to solve by someone. I just have to solve it because it's a part of my project. If it's impossible to solve analytically, tell me how to solve it numerically.
 A: First notice that $a^{\frac{1}{b}} = \Bigg(1 + \frac{b}{x}\Bigg)^{\frac{x}{b}} $. Now 
$$\begin{align}\ln a = \ln\Bigg(1  +\frac{b}{x}\Bigg)^x &\Rightarrow \Bigg (1 + \frac{b}{x}\Bigg)^{\Big(1 + \frac{x}{b}\Big)}\ln a = \Bigg (1 + \frac{b}{x}\Bigg)^{\Big(1 + \frac{x}{b}\Big)} \ln\Bigg(1  +\frac{b}{x}\Bigg)^x\\&\Rightarrow \Bigg[\Bigg (1 + \frac{b}{x}\Bigg)^{\frac{x}{b}}\Bigg]^{\Big(1 + \frac{b}{x}\Big)}\ln a = \Bigg (1 + \frac{b}{x}\Bigg)^{\Big(1 + \frac{x}{b}\Big)} \ln\Bigg(1  +\frac{b}{x}\Bigg)^x\\&\Rightarrow e^{\frac{x}{b}\ln\Big(1 + \frac{b}{x}\Big)\Big(1 + \frac{b}{x}\Big)}\ln a = \Bigg(1 + \frac{b}{x}\Bigg)^{\frac{x}{b}}\Bigg(1 + \frac{b}{x}\Bigg)\ln a\\&\Rightarrow\ln a = a^{\frac{1}{b}}\ln a \Bigg(1 + \frac{b}{x}\Bigg)e^{-\frac{x}{b}\ln\Big(1 + \frac{b}{x}\Big)\Big(1 + \frac{b}{x}\Big)}\\&\Rightarrow\ln a = a^{\frac{1}{b}}\ln a \Bigg(1 + \frac{b}{x}\Bigg)e^{-\frac{1}{b}\ln a\Big(1 + \frac{b}{x}\Big)}\\&\Rightarrow -\frac{\ln a}{b\ \  a^{\frac{1}{b}}} = -\frac{1}{b}\ln a \Bigg(1 + \frac{b}{x}\Bigg)e^{-\frac{1}{b}\ln a\Big(1 + \frac{b}{x}\Big)}\\& \Rightarrow W\Bigg(-\frac{\ln a}{b\ \  a^{\frac{1}{b}}}\Bigg) = -\frac{1}{b}\ln a \Bigg(1 + \frac{b}{x}\Bigg)\\&\Rightarrow -\frac{1}{b}\ln a  - \frac{1}{x}\ln a = W\Bigg(-\frac{\ln a}{b\ \  a^{\frac{1}{b}}}\Bigg)  \\&\Rightarrow xW\Bigg(-\frac{\ln a}{b\ \  a^{\frac{1}{b}}}\Bigg) + \frac{x}{b}\ln a = - \ln a \\&\Rightarrow x = -\frac{\ln a}{W\Bigg(-\frac{\ln a}{b\ \  a^{\frac{1}{b}}}\Bigg) + \frac{1}{b}\ln a} \\& \Rightarrow x = -\frac{b\ln a}{bW\Bigg(-\frac{\ln a}{b\ \  a^{\frac{1}{b}}}\Bigg) + \ln a}\end{align}$$
Where $W$ is the Lambert W-Function.
