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!

Let $s=\ln(a)$ and $t=\ln(a^b)$
Then $a=e^s$ and $a^b=e^t$.
But $a^b=(e^s)^b=e^{sb}$
So $e^t=e^{sb}$
Thus $t=sb$.
That is, $\ln (a^b)=b\ln(a)$.
