How to find solution for this equation

Question

I have following equation$$x\ln(1+\frac{b}{x})=b$$

where $b>0$. How to find the solution for $x$. I know how to solve equation involving $x\ln(x)$ but I don't know how to solve equation where inverse of $x$ is present.

Any help in this regard will be much appreciated.

BR

Frank

Answer 1

$$\ln(1+u)=u\implies u=0.$$ You can simply show that $u\longmapsto \ln(1+u)-u$ is strictly decreasing on $[0,\infty [$ to have the unicity.

Answer 2

Transform

$$\ln\left(1+\frac{b}{x}\right)=\frac{b}{x},$$ $$\ln\left(1+\frac{b}{x}\right)-\left(1+\frac{b}{x}\right)=1,$$

$$-\left(1+\frac{b}{x}\right)e^{-\left(1+\frac{b}{x}\right)}=-e,$$

$$-\left(1+\frac{b}{x}\right)=W(-e).$$

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Looking for a complex solution, let

$$\ln(z)=z-1,$$ or in polar coordinates

$$\ln(r)+i\theta=r\cos(\theta)+ir\sin(\theta)-1,$$

giving

$$r=\frac\theta{\sin(\theta)},\\ \ln\left(\frac\theta{\sin(\theta)}\right)=\theta\cot(\theta)-1.$$

The first nontrivial solution is found numerically as

$$\theta=1.7881880413836\cdots,\\r=1.8312905141248\cdots,\\z=-0.3949790827072\cdots+i 1.7881880413836\cdots$$