Linear Operator on General Logarithm

Theorem

Let $\phi: \R^\R \to \R^\R, y \mapsto \phi\left({y}\right)$ be a linear operator on the space of functions from $\R\to\R$.

Let $y$ be a real function such that $\forall x \in \R$, $y\left({x}\right) > \mathbf{0}\left({x}\right) = 0$.

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

Then:

$\phi \left({\log_a y}\right) = \dfrac 1 {\ln a} \left({\phi \left({\ln y}\right)}\right)$

where $\ln$ is the natural logarithm.

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \phi \left({\log_a y}\right)\)	\(=\)	\(\displaystyle \phi \left({\frac {\ln y}{\ln a} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Change of Base of Logarithm
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {\ln a} \left({\phi \left({\ln y}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	definition of linear operator (as $\frac 1 {\ln a}$ is constant)

$\blacksquare$