# Invert big-O involving logarithms while retaining a good error term

I have an equation $$y=\frac{kx}{\log(y/k)-1}+O(1)$$

which I would like to solve for $y$ in terms of $x$ ($k$ is constant). Clearly $y\sim kx/\log x$ but I would like to preserve the error term. Can this be done? I can iterate a few times but that seems only to get an error of the form $O\left(\frac{x}{\log^cx}\right)$ which is quite a bit worse than the original error.

Iterating gives $$\begin{array}{rl} y=&\frac{kx}{\log\left(\frac{x}{\log(y/k)-1}+O(1)\right)-1}+O(1)\\ =&\frac{kx}{\log\left(\frac{x}{\log(y/k)-1}\right)-1}+O(1/\log x)+O(1)\\ =&\frac{kx}{\log\left(\frac{x}{\log(y/k)-1}\right)-1}+O(1)\\ =&\frac{kx}{\log\log x-\log\log y-\log\frac{1+\log k}{\log y}+O\left(\frac{1}{\log^2y}\right)-1}+O(1)\\ =&\frac{kx}{\log\log x-\log\log y-\log\frac{1+\log k}{\log y}-1}+O\left(\frac{x}{\log^4x}\right)\\ \end{array}$$ if I am not mistaken. Of course it can be repeated (substituting, say, the last line into the third line) but the error seems to stay in the same form.

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