#1**+2 **

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

CPhill
May 16, 2017

#2**+2 **

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

hectictar
May 16, 2017

#3**+1 **

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!!!!!

CPhill
May 17, 2017