Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x

Theorem
Let $b \in \R$ be a real number.


 * $\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor \iff b \in \Z_{> 1}$

where $\left\lfloor{x}\right\rfloor$ denotes the floor of $x$.

Necessary Condition
Let:
 * $\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor$

Let $x = b$.

Then:
 * $\left\lfloor{\log_b b}\right\rfloor = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$

$b \notin \Z$.

Thus by Proof by Contradiction:
 * $b \in \Z$

But for $\log_b$ to be defined, $b > 0$ and $b \ne 1$.

Hence:
 * $b \in \Z_{> 1}$

Sufficient Condition
Let $b \in \Z_{> 1}$.

Let $\left\lfloor{\log_b x}\right\rfloor = n$.

Then: