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# can someone chalenge me with a logarithm question please

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can someone chalenge me with a logarithm question please

Guest May 29, 2017
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#1
+4716
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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

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

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

hectictar  May 29, 2017
#2
+76925
+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

CPhill  May 29, 2017

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