How to solve this equation involving natural log

Question

I have $2$ related questions:

1. I know value of $\ln(x) / \ln(y)$, say it is $v$, how can I find value of $x/y$?

2. If $\ln(x) = v_1$ and $\ln(y) = v_2$ , what is $x/y$ ?

Thanks for your help. Apologies if these are very basic questions.

Answer 1

Note $ln(x) =v_1\implies x= e^{v_1}$ similarly $y= e^{v_2}$

Thus $x\over y$ $ = \frac{e^{v_1}}{e^{v_2}}=e^{{v_1}-{v_2}}$

Answer 2

1. You can't. Note that $\dfrac{\log(3^2)}{\log(3)}=\dfrac{\log(2^2)}{\log(2)}=2$, but $\dfrac{3^2}3\ne\dfrac{2^2}2$.
2. $\dfrac xy=\dfrac{e^{v_1}}{e^{v_2}}$

Answer 3

$\frac{\ln x}{\ln y}=v$ implies $\ln x=v \ln y= \ln y^v,$ hence $x=y^v, $ thus $x/y= y^{v-1}.$