Hey!! I might have figured out a way!!!
\(3\log(x-2)=\log(2x)-3 \\~\\ \log(x-2)=\frac{\log(2x)-3}{3}\)
And...
\(\log_{10}(n)=a \quad \rightarrow \quad 10^a=n\)
That means:
\(10^{\frac{\log(2x)-3}{3}}=(x-2) \\~\\ (10^{\frac{\log(2x)-3}{3}})^3=(x-2)^3 \\~\\ 10^{\log(2x)-3}=(x-2)^3 \\~\\ 10^{\log(2x)}*10^{-3}=(x-2)^3 \\~\\ \frac{2x}{10^3}=x^3-6x^2+12x-8 \\~\\ 0=x^3-6x^2+(12-\frac{2}{10^3})x-8 \\~\\ 0=x^3-6x^2+11.998x-8\)
Annnd then....I dunno how to solve that..I thought once I got rid of the logs it would be easy....
But I checked this on WolframAlpha:
http://www.wolframalpha.com/input/?i=0%3Dx%5E3-6x%5E2%2B11.998x-8
and it said x ≈ 2.16294