I have read this thread:
Expressing logarithms as ratios of natural logarithms
It says this and I quote: " Consider $y=\log_b x$. Then, by definition, $b^y=x$ and so $y \ln b = \ln x$. Thus,
$$\log_b x=\frac{\ln x}{\ln b}$$ "
Could somebody please expand on this? I know how logarithms work in general, yet why the statement
$y * \ln b = \ln x$
follows from
$b^y=x$
is beyond me. How does the natural logarithm play into that? I have only known the role of the natural base in the realm of continous growth or decay.
Best Answer
If one knows that $$ \ln (x^y)=y\ln x,\qquad x>0,\,y\in\mathbb{R}, \tag1 $$ then applying $$ A=B \implies \ln A = \ln B,\qquad A>0,\,B>0, \tag2 $$ to $$ b^y=x $$ gives $$ \ln (b^y)=\ln x $$ and using $(1)$ yields $$ y \ln b=\ln x $$ as desired.