Exponentiation of log of floor and ceil

Dependencies:

  1. Floor and ceil of log

Let $a, b \in \mathbb{Z}$ where $b \ge 2$ and $a \ge 1$. Then \[ \frac{a+1}{b} \le b^{\left\lfloor \log_b a \right\rfloor} \le a \] If $a \ge 2$, then \[ a \le b^{\left\lceil \log_b a \right\rceil} \le b(a-1) \]

Proof

\begin{align} & \log_b(a+1) - 1 \le \left\lceil \log_b (a+1) \right\rceil - 1 = \left\lfloor \log_b a \right\rfloor \le \log_b a \\ &\implies \frac{a+1}{b} \le b^{\left\lfloor \log_b a \right\rfloor} \le a \end{align}

When $a \ge 2$, \begin{align} & \log_b a \le \left\lfloor \log_b a \right\rfloor = \left\lceil \log_b (a-1) \right\rceil + 1 \le \log_b (a-1) + 1 \\ &\implies a \le b^{\left\lceil \log_b a \right\rceil} \le b(a-1) \end{align}

Dependency for: None

Info:

Transitive dependencies:

  1. Floor and ceil of log