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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

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  • $\begingroup$ First of all, divide it by $x$ and denote $b/x$ by a new variable. $\endgroup$ – Andrew Miloradovsky Mar 3 '16 at 8:26
  • $\begingroup$ yes i did it but then i get inverse at the outside of logrithm $\endgroup$ – Frank Moses Mar 3 '16 at 8:27
  • $\begingroup$ @AndrewMiloradovsky BTW what will be the next step can you elaborate it a bit. Thanks $\endgroup$ – Frank Moses Mar 3 '16 at 8:29
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    $\begingroup$ I do not believe there is a real valued solution for $x$ if you take $b > 0$. $\endgroup$ – Areyouheaty Mar 3 '16 at 8:32
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    $\begingroup$ $\ln {(1+t)} \le t$, when $t=0$ get "=". so you don't have solution for $x$ unless $b=0$ $\endgroup$ – chenbai Mar 3 '16 at 8:33
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$$\ln(1+u)=u\implies u=0.$$ You can simply show that $u\longmapsto \ln(1+u)-u$ is strictly decreasing on $[0,\infty [$ to have the unicity.

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Transform

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

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

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


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

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

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  • $\begingroup$ So, no real solutions, I guess. $\endgroup$ – Claude Leibovici Mar 3 '16 at 9:55
  • $\begingroup$ @ClaudeLeibovici: quite right. $\endgroup$ – Yves Daoust Mar 3 '16 at 10:30

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