Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$

Question

Let $f : \mathbb N \to \mathbb N$ be a function such that $f(n) \le 1 + \log_b n$ for some base $b$ and all $n$. Now let $n \in \mathbb N$ have the property that $$ \frac{r^m - 1}{r-1} \le n < \frac{r^{m+1}-1}{r-1} $$ for some $m \in \mathbb N$ and $r > 1$ with $r \in \mathbb N$. Does it follow that $f(n) \le \lceil m \cdot \log_b r \rceil$?

Answer 1

No, it does not follow.

Let $b=2$, $r=1.1$, and $m=1$. Let's assume that $f(n) = 1+\lfloor \log_2(n) \rfloor$.

Then $n$ must satisfy $$ 1 \leq n < r+1 = 2.1.$$ So let $n=2$. Then $f(2) = 1 + \log_2(2) = 2$, but the proposed upper bound is $\lceil m \cdot \log_b r \rceil = \lceil \log_2(1.1) \rceil = 1$.