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

??

 

how do you have find x

Guest May 16, 2017
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5+0 Answers

 #1
avatar+77022 
+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

 

 

 

cool cool cool

CPhill  May 16, 2017
 #2
avatar+4734 
+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.... sad

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
avatar+77022 
+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!!!!!

 

 

cool cool cool

CPhill  May 17, 2017
edited by CPhill  May 17, 2017
 #4
avatar
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
avatar+26242 
+1

Here's a numerical approach:

 

.So x ≈ 2.162939

.

Alan  May 17, 2017

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