This is not quite an answer, but also way too big for a comment.
Euler made a powerful observation, that $a_0+a_0a_1+a_0a_1a_2+\dots+a_0a_1a_2\dots a_n$ equals
$$\cfrac{a_0}{1-
\cfrac{a_1}{1+a_1-
\cfrac{a_2}{1+a_2-
\cfrac{\ddots}{\ddots-
\cfrac{a_{n-1}}{1+a_{n-1}-
\cfrac{a_n}{1+a_n}}}}}}$$
Using this and taking $n\to\infty$, one can show that if $x=1+\sum_{i=1}^{\infty}\left(\prod_{j=1}^ia_i\right)$ then
$$x=\cfrac{1}{1-
\cfrac{a_1}{1+a_1-
\cfrac{a_2}{1+a_2-
\cfrac{a_3}{1+a_3-\ddots}}}}$$
This, together with an application of an equivalence transformation, is what's behind Euler's celebrated continued fraction representation for $e^z$ $(z>0)$:
$$e^z=\cfrac{1}{1-\cfrac{z}{1+z-\cfrac{z}{2+z-\cfrac{2z}{3+z-\cfrac{3z}{4+z-\ddots}}}}}$$
The expression for $\log\left(\frac{1+z}{1-z}\right)$ comes from the power series
$$\log\left(\frac{1+z}{1-z}\right)=2\sum_{n=0}^\infty\frac{z^{2n+1}}{2n+1},$$
using the same principle as above.
Indeed, we can write
\begin{align}
\log\left(\frac{1+z}{1-z}\right)
&=2z\left(1+\frac{z^2}3+\frac{z^4}5+\dots\right)\\
&=2z\left(1+\frac{z^2}3+\frac{z^2}{3}\frac{z^2}{5/3}+\dots\right)
\end{align}
and apply an equivalence transformation at the end to tidy things up.
The continued fraction representation for $\log(1+z)$ is a bit more complicated.
It comes from Gauss's continued fraction representations for hypergeometric functions.
Indeed, for $|z|<1$ we have
\begin{align}
\log(1+z)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}nz^n
&=z\sum_{m=0}^\infty \frac{(-1)^m}{m+1}z^m\\
&=z\sum_{m=0}^\infty\frac{(-z)^m}{m!}\frac{(1)_m(1)_m}{(2)_m}\\
&=z\,\,{}_2F_1(1,1,;2;-z)
\end{align}
This yields a continued fraction representation for $\log(1+z)$ that is still valid outside of the branch cut, ie, outside $(-\infty,-1)$.
If you don't mind $\log b$ showing up, there's an obvious formula.
Taking $a^*=a-1$ we get that $\frac{\log a
}{\log b}=\frac1{\log b} \log(1+a^*)$ must equal
\begin{align}
\frac1{\log b}\cdot
&\cfrac{a^*}{1+
\cfrac{a^*}{2+
\cfrac{a^*}{3+
\cfrac{4a^*}{4+
\cfrac{4a^*}{5+
\cfrac{9a^*}{6+
\cfrac{9a^*}{7+\ddots}}}}}}}\\
=&\cfrac{a^*}{\log b+
\cfrac{a^*\log b}{2+
\cfrac{a^*}{3+
\cfrac{4a^*}{4+
\cfrac{4a^*}{5+
\cfrac{9a^*}{6+
\cfrac{9a^*}{7+\ddots}}}}}}}
\end{align}
If you're not happy with $\log b$ showing up near $a^*$, you can instead write
\begin{align}
&\cfrac{a^*}{\log b+\frac1{\frac1{\log b}}\cdot
\cfrac{a^*}{2+
\cfrac{a^*}{3+
\cfrac{4a^*}{4+
\cfrac{4a^*}{5+
\cfrac{9a^*}{6+
\cfrac{9a^*}{7+\ddots}}}}}}}\\
= &\cfrac{a^*}{\log b+
\cfrac{a^*}{\frac{2}{\log b}+
\cfrac{a^*}{3\log b+\frac1{\frac1{\log b}}\cdot
\cfrac{4a^*}{4+
\cfrac{4a^*}{5+
\cfrac{9a^*}{6+
\cfrac{9a^*}{7+\ddots}}}}}}}
\end{align}
Proceeding in this manner, we get
$$\cfrac{a^*}{\log b+
\cfrac{a^*}{\frac{2}{\log b}+
\cfrac{a^*}{3\log b+
\cfrac{4a^*}{\frac4{\log b}+
\cfrac{4a^*}{5\log b+
\cfrac{9a^*}{\frac6{\log b}+
\cfrac{9a^*}{7\log b+\ddots}}}}}}}$$
$$\frac{\log a}{\log b}=\log_ba,$$
I think you will find this link somewhat interesting.
– Fimpellizzeri Oct 26 '17 at 19:31