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# 3log(x-2)=log(2x) - 3

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3log(x-2)=log(2x) - 3

??

how do you have find x

Guest May 16, 2017
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#1
+80866
+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
+5888
+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
+80866
+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
edited by CPhill  May 17, 2017
#4
0

yep, i realised after your comment that the question was a graohic calculator question!

thank you so much for your help anyways, gave me a real stump.

Guest May 17, 2017
#5
+26397
+1

Here's a numerical approach:

.So x ≈ 2.162939

.

Alan  May 17, 2017

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