Linear Operator on General Logarithm

Theorem

Let $\phi: \R^\R \to \R^\R, y \mapsto \map \phi y$ be a linear operator on the space of functions from $\R\to\R$.

Let $y$ be a real function such that:

$\forall x \in \R: \map y x > \map \bszero x = 0$.

Let $\log_a y$ be the logarithm of $y$ to base $a$.

Then:

$\map \phi {\log_a y} = \dfrac 1 {\ln a} \paren {\map \phi {\ln y} }$

where $\ln$ is the natural logarithm.

Proof

\(\ds \map \phi {\log_a y}\)

\(=\)

\(\ds \map \phi {\frac {\ln y}{\ln a} }\)

Change of Base of Logarithm

\(\ds \)

\(=\)

\(\ds \frac 1 {\ln a} \paren {\map \phi {\ln y} }\)

Definition of Linear Operator

$\blacksquare$