Floor of log equation $S=\left(\lfloor\log_{10}(x)\rfloor+1\right)x - \frac{10^{\lfloor\log_{10}(x)\rfloor+1}-10}{9}$ I must find 'x' and I don't know how to solve the following equation.
Does it have a solution? How can I solve it?
$$
S=\left(\lfloor\log_{10}(x)\rfloor+1\right)x - \frac{10^{\lfloor\log_{10}(x)\rfloor+1}-10}{9}
$$
$$S,x\in\mathbb N$$

For an integer $k\ge 0$, if $10^k \le x < 10^{k+1}$, then $\lfloor \log_{10}x\rfloor = k$ so
$$
S(x) = (k+1)x -\frac{10^{k+1}-10}{9}=(k+1)x-\frac{10}{9}(10^k-1)
$$
So, for example, if $1\le x \le 9$, then $S(x) = x$. If $10\le x \le 99$, then $S=2x-10$. If $100\le x\le 999$, then $S=3x-110$, and so on.
The first few $S$ values are
$$
\begin{array}
&x = & 1 & 2 &\dots & 9 & 10 & 11 & 12 & \dots & 99 & 100 & 101 & 102 & \dots 999\\ 
S = & 1 & 2 &\dots & 9 & 10 & 12 & 14 & \dots & 188 & 190 & 193 & 196 & \dots 2887\\
\end{array}
$$
