3 log ( x - 2) = log (2x) - 3
Don't see a way to solve this algebraically [but maybe someone else on here knows one ]......
Here is a graphical solution : https://www.desmos.com/calculator/tbdqhaxixk
The solution is x ≈ 2.163
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
Good job, hectitctar......at least you got it down to a cubic ....
But you're correct.....it's still hard to solve....!!!!
Note.... There is a "formula" for solving a cubic, but it's pretty messy !!!
Also......here's a procedure involving some substitutions.....
http://www.sosmath.com/algebra/factor/fac11/fac11.html
Not sure it's worth pursuing for this problem, but if you can work it out.....I'd be willing to chip in a point (or two).......LOL!!!!!