# Rewriting logarithms as the ratios of natural logarithms

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.

• Possible duplicate of Confusion regarding the Logarithmic function change of base formula Commented Aug 9, 2016 at 20:03
• Well, given the equation $b^y = x$, we can take the natural logarithmm $\ln$ of both sides to get $\ln(b^y)=\ln(x)$, assuming $x$ and $b$ are positive real numbers. Then you simply apply one of the rules of logarithms $\log(a^b)=a\cdot \log(b)$. Therefore, $y\cdot \ln(b)=\ln(x)$. Commented Aug 9, 2016 at 20:06
• It comes from the Change of Base rule. The choice of $\ln$ is arbitrary, but presumably convenient in this case.
– user301988
Commented Feb 10, 2017 at 21:39

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.