Find a limit without l'Hospital: $\lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$

If we use the well known limit

$$\lim_{X\to 0}\frac{\ln(1+X)}{X}=1$$

with $X=2^x\;$ and $\;X=3^x,\;\;$we find

$$\lim_{x\to-\infty}\left( \frac{2}{3} \right)^x=+\infty$$