Note that [tex]\log_{10}1=0[/tex], [tex]\log_{10}10=1[/tex], [tex]\log_{10}100=2[/tex], and so on. The function [tex]\log_{10}x[/tex] is continuous and increasing for all [tex]x>0[/tex], so when [tex]1<x<10[/tex], we have [tex]0<\log_{10}x<1[/tex]; when [tex]10<x<100[/tex], [tex]1<\log_{10}x<2[/tex]; and so on.
This means
[tex]\lfloor\log_{10}x\rfloor=\begin{cases}0&\text{for }1\le x<10\\1&\text{for }10\le x<100\\2&\text{for }100\le x<1000\\\vdots\end{cases}[/tex]
which means we can capture the number of digits of [tex]n\in\mathbb N[/tex] with the function [tex]\lfloor\log_{10}n\rfloor+1[/tex].
So the problem is the same as finding positive integer solutions to
[tex]\lfloor\log_{10}n\rfloor+1=n[/tex]
We know that [tex]n=1[/tex] has one digit, so clearly this must be a solution. We need to show that this is the only solution.
Recall that [tex]\dfrac{\mathrm d}{\mathrm dx}[\log_{10}x+1]=\dfrac1{\ln10\,x}[/tex], while [tex]\dfrac{\mathrm d}{\mathrm dx}[x]=1[/tex]. This means [tex]\log_{10}x[/tex] increases at a much slower rate than [tex]x[/tex] as [tex]x\to\infty[/tex]. We know the two functions intersect when [tex]x=1[/tex]. Therefore it's clear that [tex]x>\log_{10}x+1[/tex] for all [tex]x>1[/tex].
Now, it's always the case that [tex]\lfloor f(x)\rfloor\le f(x)[/tex], so we're essentially done:
[tex]x>\log_{10}x+1\ge\lfloor\log_{10}x\rfloor+1[/tex]
which means there are no other solutions than [tex]n=1[/tex].
