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Log4 x - log4 (x-3) = 5

  1. Jul 6, 2011 #1
    1. The problem statement, all variables and given/known data

    log4 x - log4 (x-3) = 5

    2. Relevant equations



    3. The attempt at a solution

    log4 x/x-3 = 5
     
    Last edited by a moderator: May 12, 2014
  2. jcsd
  3. Jul 6, 2011 #2

    dextercioby

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    Re: logarithm

    OK, so it's something like

    [tex] \log_{4} \frac{x}{x-3} = 5 [/tex]

    What do you do next ?
     
  4. Jul 6, 2011 #3
    Re: logarithm

    i think

    x/ x-3 = 4^5
     
  5. Jul 6, 2011 #4

    dextercioby

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    Re: logarithm

    Good thinking. Next ?
     
  6. Jul 6, 2011 #5
    Re: logarithm

    x[itex]/[/itex] x-3 = 1024
     
  7. Jul 6, 2011 #6

    dextercioby

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    Re: logarithm

    So it's very easy to find the x, right ?
     
  8. Jul 6, 2011 #7
    Re: logarithm

    no thats where i am stuck
     
  9. Jul 6, 2011 #8
    Re: logarithm

    i would probably do this:

    x = 1024 (x-3)
     
  10. Jul 6, 2011 #9
    Re: logarithm

    :confused:??
     
  11. Jul 6, 2011 #10

    dextercioby

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    Re: logarithm

    So it's like 3072 = 1023 x. Do you agree ?
     
  12. Jul 6, 2011 #11
    Re: logarithm

    yes now it will be
    x = 1023x - 3072
     
  13. Jul 6, 2011 #12

    dextercioby

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    Re: logarithm

    I think you're missing something very simple. You've got an equation in which you must what numerical value hides under 'x'.

    So [itex] x= 1024\cdot (x-3) = 1024\cdot x - 1024\cdot 3 = 1024\cdot x - 3072 [/itex]

    Now subtract x from both terms of [itex] x= 1024\cdot x - 3072 [/itex].

    What do you get ?
     
  14. Jul 7, 2011 #13
    Re: logarithm

    ok, so if i am going to subtract x from both terms, i would end up with:

    x = 1024 - 3072
     
  15. Jul 7, 2011 #14
    Re: logarithm

    x = 1024 - 3072
     
  16. Jul 7, 2011 #15

    Mentallic

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    Re: logarithm

    What?

    Don't you know basic algebra? Solve for x in 3x=x+2, now apply the same idea to solve for x in x=1024x-3072
     
  17. Jul 7, 2011 #16
    Re: logarithm

    as u can see i am not good at this, ok here goes

    1024x-x=3072
    1023x = 3072
     
  18. Jul 7, 2011 #17

    Mentallic

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    Re: logarithm

    Well then you need to go back and cover basic algebra again. You can't afford to lose that many marks on logarithms just because you don't know your algebra.

    Yes and now? If I asked you to solve for x why haven't you given us x=... ?
     
  19. Jul 7, 2011 #18
    Re: logarithm

    x = 3072-1023
    x = 2049
     
  20. Jul 7, 2011 #19

    Mentallic

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    Re: logarithm

    No, it's not 1023+x=3072, it's 1023x=3072. You really need to go back and catch up on what you've missed out on.
     
  21. Jul 7, 2011 #20
    Re: logarithm

    so i would divide both sides by 1023

    1023x = 3072

    1023x/1023 = 3072/1023
     
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