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can someone chalenge me with a logarithm question please

 May 29, 2017
 #1
avatar+7347 
+1

OOOH ! I got one...it requires more than just logarithm knowledge though!

 

Be warned! I spent a VERY LONG time on this and never got the answer crying

 

But....you asked for it....here it is:

 

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

 May 29, 2017
 #2
avatar+94235 
+2

 

3log ( x - 2)   =  log (2x) - 3

 

Note that we can write 3   as  log (1000)

 

So  we have

 

3 log (x - 2)  =  log (2x)  - log (1000)    and we can write

 

log ( x - 2)^3  =  log  [ (2x) / 1000]

 

log ( x - 2)^3  =  log (x / 500)       we can forget the logs and solve for

 

(x - 2)^3  =  ( x / 500)      simplify

 

x^3 - 6x^2 + 12x - 8  =  x / 500       multiply both sides by 500

 

500x^3  - 3000x^2 + 6000x - 4000  = x      subtract x from both sides

 

500x^3 - 3000x^2 + 5999x - 4000  =  0     difficult to solve this, algebraically....

 

WolframAlpha shows the real solution as

 

x ≈  2.1629

 

 

 

 

cool cool cool

 May 29, 2017

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