Linear Operator on General Logarithm

Theorem
Let $\phi: \R \to \R$ be a linear operator on the reals.

Let $y$ be a real function such that $\phi\left({y}\right) > 0$.

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.