Can all logarithm problems be solved algebraically? Trying to solve $\log_2(x-1)=\log_3(x+1)$ and can't seem to get it algebraically.  Tried changing bases, moving things around, but can't seem to crack it.
 A: $$
\log_2(x-1)=\log_3(x+1)
$$
$$
\log_2(x-1) = \frac{\log_2 (x+1)}{\log_2 3}
$$
$$
(\log_2 3)(\log_2(x-1)) = \log_2 (x+1)
$$
$$
\log_2((x-1)^{\log_2 3}) = \log_2 (x+1)
$$
$$
(x-1)^{\log_2 3} = x+1
$$
At this point I might apply Newton's method.
Pedja's earier answer did everything right until a mistake near the end, but I prefer not to bring in $e$ when the number $e$ is not essential to the situation.  Avoiding $e$ was really the reason why I felt this is worth answering.
Later note: Pedja has fixed the error.
A: $$\log_2 (x-1)=\log_3(x+1) \Rightarrow \frac{\ln (x-1)}{\ln 2}=\frac{\ln (x+1)}{\ln 3} \Rightarrow \ln 3 \cdot \ln (x-1)= \ln 2 \cdot \ln (x+1) \Rightarrow$$
$$\Rightarrow \ln (x-1)^{\ln 3}-\ln (x+1)^{\ln 2}=0 \Rightarrow \frac{(x-1)^{\ln 3}}{{(x+1)^{\ln 2}}}=1 \Rightarrow (x-1)^{\ln 3}=(x+1)^{\ln 2} \Rightarrow$$
$$\Rightarrow (x-1)=(x+1)^{\log_3 2} \Rightarrow (x+1)-2=(x+1)^{\log_3 2}$$
If we make substitution $~u=(x+1)~$ we get :
$u-u^{\log_3 2}=2$
According to WolframAlpha this equation can be solved using numerical methods , so:
$u \approx 4.6298 \Rightarrow x \approx 3.6298$
