how to solve a $\log(1+\frac1x)*x$ function

Question

I know lambert function is available to solve function like $xln(x)$, I wonder if there is a similar way I can solve function $b - x log_2(1+\frac{a}{x})=0$.

Answer 1

replace $x=1/(t-1)$ the equation becomes $$\frac1{t-1}\log t=0$$ The first term isn't equal to $0$ anywhere hence $\log t=0$ is the only solution and this gives $t=1$ at which point the function isn't defined. Hence there are no solutions to this.

For your second question $$b\ln 2 = t\ln(1+\frac at)$$ Now substitutue $x=a/(t-1)$ $$b\ln2 =k=\frac{a}{x-1}\ln x \\ xe^{-\frac ka x}= e^{-\frac ka} \\ \frac{-k}axe^{\frac{-k}ax}=\frac{-k}ae^{\frac{-k}a} \\ W(\frac{-k}ae^{\frac{-k}a})=\frac{-k}a x \\ x=\frac{W(\frac{-k}ae^{\frac{-k}a})}{\frac{-k}{a}}$$ Where $W$ is the lambert function and $ k=\ln 2$

Answer 2

You want to solve for $x$ the equation $$x \log_2\left(1+\frac{a}{x}\right)=b$$ Using natural logarithms, dividing each side by $a$ and defining $t=\frac a x$, we ned with $$\frac{\log (t+1)}{t}=k \qquad \text{where} \qquad k=\frac{b }{a}\log (2)$$ The lhs is a continuous function and provided that $k>0$ (that is to to say $a\,b>0$) there is only one root in the reak domain.

It just write $$t=-1-\frac{1}{k}W\left(-k\,e^{-k} \right)$$