# 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.

$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{> 1}$

where $\floor x$ denotes the floor of $x$.

## Proof 1

### Necessary Condition

Let:

$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x}$

Let $x = b$.

Then:

$\floor {\log_b b} = \floor {\log_b \floor b}$
 $\ds \floor {\log_b b}$ $=$ $\ds \floor {\log_b \floor b}$ $\ds \leadsto \ \$ $\ds \floor 1$ $=$ $\ds \floor {\log_b \floor b}$ Logarithm of Base $\ds \leadsto \ \$ $\ds 1$ $=$ $\ds \floor {\log_b \floor b}$ Real Number is Integer iff equals Floor

Aiming for a contradiction, suppose $b \notin \Z$.

 $\ds \floor b$ $<$ $\ds b$ Floor of Non-Integer $\ds \leadsto \ \$ $\ds \log_b \floor b$ $<$ $\ds 1$ $\ds \leadsto \ \$ $\ds \floor {\log_b \floor b}$ $<$ $\ds \floor {\log_b b}$ $\ds \leadsto \ \$ $\ds \floor {\log_b \floor b}$ $\ne$ $\ds \floor {\log_b b}$ which contradicts $\floor {\log_b b} = \floor {\log_b \floor b}$

$b \in \Z$

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

Hence:

$b \in \Z_{> 1}$

$\Box$

### Sufficient Condition

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

Let $\floor {\log_b x} = n$.

Then:

 $\ds n$ $=$ $\, \ds \floor {\log_b x} \,$ $\ds$ $\ds \leadstoandfrom \ \$ $\ds n$ $\le$ $\, \ds \log_b x \,$ $\, \ds < \,$ $\ds n + 1$ Integer equals Floor iff Number between Integer and One More $\ds \leadstoandfrom \ \$ $\ds b^n$ $\le$ $\, \ds x \,$ $\, \ds < \,$ $\ds b^{n + 1}$ Power Function is Strictly Increasing over Positive Reals $\ds \leadstoandfrom \ \$ $\ds b^n$ $\le$ $\, \ds \floor x \,$ $\, \ds < \,$ $\ds b^{n + 1}$ Number not less than Integer iff Floor not less than Integer $\ds \leadstoandfrom \ \$ $\ds n$ $\le$ $\, \ds \log_b \floor x \,$ $\, \ds < \,$ $\ds n + 1$ Power Function is Strictly Increasing over Positive Reals $\ds \leadstoandfrom \ \$ $\ds n$ $=$ $\, \ds \floor {\log_b \floor x} \,$ $\ds$ Integer equals Floor iff Number between Integer and One More $\ds \leadstoandfrom \ \$ $\ds \floor {\log_b x}$ $=$ $\, \ds \floor {\log_b \floor x} \,$ $\ds$ Definition of $n$

$\blacksquare$

## Proof 2

We have that:

Logarithm is Strictly Increasing

and:

Real Natural Logarithm Function is Continuous

Suppose that $\log_b x \in \Z$: let $\log_b x = n$, say.

Then:

$x = b^n$

It follows that:

$x \in \Z \iff b \in \Z$
$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{>1}$

$\blacksquare$