ELF.AP.Proof.4 : $\ln(y) = \log_{base}( y ) \cdot ln(base)$ or $\log_{base}(y) = \frac{\ln(y)} {\ln(base)}$
Proof : Apply $\ln(b^p)=p\ln(b)$ with $b = base ; p = \log_{base}(y)$.
So $y = b^p = base^{ \log_{base}(y)}$ and thus $\ln(y) = \log_{base}(y) \cdot \ln(base).$

This can also be proven directly by using the inverse relation of  $\ln$ and  $\exp_{e}$ for the $base$.
Let $\ln(base)=a$  so $e^a=base$.
Here's what that looks like on a mapping diagram: Next we consider  $x= \log_{base} (y)$ so  $base^x = y$.
Then  $(e^a)^x =e^{ax} = y$ and thus we have the corresponding logarithmic equation:
$\ln(y)=ax= \ln(base) \cdot \log_{base}(y)$.

Here's what that looks like on a mapping diagram: And here's a mapping diagram showing the product of the logarithm visualized as well: Finally here is a dynamic visualization of the proof using mapping diagrams:
 Proof of $\ ln(y) = \log_{base}( y ) \cdot ln(base)$ This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Martin Flashman, 26 Oct 2014, Created with GeoGebra