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?
$$
\log(x) = \log(100x) \, \log(2)
$$
 A: 
How would I then find the exact solution of this problem?

Manipulate the equation to isolate $x$.
\begin{align*}
\log(x) &= (\log(100)+\log(x))\log(2)  \\
\log(x) &=\log(100)\log(2)+\log(x)\log(2)\\
\log(x)-\log(x)\log(2)&=\log(100)\log(2)\\
\log(x)(1-\log(2))&=\log(100)\log(2)  \\
\log(x)&=\log(100)\log(2)/(1-\log(2))\\
\end{align*}
Then resolve $x$ with whatever base your logarithm is using. E.g. with base 10, 
$$x\approx7.267$$
A: 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)
A: Fill in details after you check the basic properties of logarithms, and assuming $\;\log=\log_{10}\;$:
$$\log x=\log 100x\cdot\log2=\left(\log100+\log x\right)\log2\implies$$
$$(1-\log2)\log x=2\log2\implies\ldots$$
A: $\log(x) = (\log(100) + \log(x))\cdot\log2$
$\log(x) = \log(100)\cdot\log(2) + \log(x)\cdot\log(2)$
This is a linear equation in $\log(x)$!
