How can I derive the rules for the natural logarithm from the rules for the exponential functions? I would like to use the calculation rules for exponential function $$1)\quad e^{x+y}=e^{x}\cdot e^{y}\quad\text{ and }\quad2)\quad(e^a)^b=e^{a\cdot b}$$ to derive the following calculation rules for the natural logarithm: $$\ln(x\cdot y)=\ln(x)+\ln(y)\quad\text{ and }\quad \ln(a^b)=b\cdot\ln(a)$$
My solution so far looks like this:
1)\begin{align*}
e^{x+y}&=e^{x}\cdot e^{y}& \ln()\text{ on both sides}\\
\ln(e^{x+y})&=\ln(e^x\cdot e^y)\\
x+y&=\ln(e^x\cdot e^y)
\end{align*}
From here, I don't know any further.
2)\begin{align*}
(e^a)^b&=e^{a\cdot b}& \ln()\text{ on both sides}\\
\ln(e^a)^b&=\ln(e^{a\cdot b})\\
\ln(e^a)^b&=a\cdot b
\end{align*}
I can't get any further here either.
I really appreciate your help and tips. Thanks in advance!

If $x=e^a$ and $y=e^b$, then\begin{align}\ln(xy)&=\ln\left(e^ae^b\right)\\&=\ln\left(e^{a+b}\right)\\&=a+b\\&=\ln(x)+\ln(y).\end{align}And\begin{align}\ln\left(a^b\right)&=\ln\left(e^{b\ln a}\right)\\&=b\ln(a).\end{align}
