Multiplying two logarithms I've searched for some answer already, but couldn't find any solution to this problem. Apparently, there's no rule for the product of two logarithms. How would I then find the exact solution of this problem?
$$
\log(x) = \log(100x) \, \log(2)
$$

The question does not specify the base $B$ of the logarithm, but it will affect the solution, so we make it explicit:
\begin{align}
\log_B(x) 
&= \log_B(100\, x) \, \log_B(2) \\
&= (\log_B(100) + \log_B(x)) \, \log_B(2) \iff \\
(1 - \log_B(2)) \log_B(x)  
&= \log_B(2) \log_B(100) \\
\end{align}
For $B = 2$ the LHS vanishes and we have no solution, as the logarithms on the RHS do not vanish.
For $B \ne 2$ we can continue:
\begin{align}
\log_B(x) = \frac{\log_B(2) \, \log_B(100)}{1 - \log_B(2)} = f(B) \iff \\
x = B^{f(B)} =  B^{(\log_B(2) \, \log_B(100))/(1 - \log_B(2))} 
\end{align}
For $B=e$ one gets
$$
x = e^{f(e)} = e^{10.4025\dotsb} = 32944.48\dotsb
$$
Here are graphs of $f(B)$:


(Links to larger versions: left, right)
