Linear Operator on General Logarithm
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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 |
$\blacksquare$