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Homework Help: Simply Logarithms problem

  1. Jul 8, 2006 #1
    1) ((logx)/(log4))^2=((logx^5)/(log4))-4

    how do you these problmes...i mean..is there anyway of simplfing the expression ...i am stuck..
    regards
    vijay
     
  2. jcsd
  3. Jul 8, 2006 #2
    Use the fact [itex]\log {a^b} = b\log{a}[/itex] and simplify.
     
  4. Jul 8, 2006 #3
    yea....i used it....but this sum is different..seriously..try it
     
  5. Jul 8, 2006 #4
    Can you show your work?
     
  6. Jul 8, 2006 #5
    mr neutrino...
    i can show my work but it is too long....
    i ll jus ask you a simple question...
    what is (logx/log4)^2...i mean...how do you simplyfy it in terms of logx^5/log4?
     
  7. Jul 8, 2006 #6

    arildno

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    First, write your equation as:
    [tex](\frac{\log(x)}{\log(4)})^{2}=(\frac{\log(4)}{\log(x^{5})})^{4}[/tex]
    Then, collect the terms on the LHS and rewrite your equation as:
    [tex](\frac{\log(x)}{\log(4)}-(\frac{\log(4)}{\log(x^{5})})^{2})*(\frac{\log(x)}{\log(4)}+(\frac{\log(4)}{\log(x^{5})})^{2})=0[/tex]
    Now, you should be able to continue a bit!
     
  8. Jul 8, 2006 #7
    Actually you've got to simplify the second term (logx^5/log4) using the rule I mentioned earlier, and then see if the resulting equation resembles something you've come across before. Remember that log4 is just a constant; don't worry too much about that part.
     
    Last edited: Jul 8, 2006
  9. Jul 8, 2006 #8
    I believe it was a -4 at the end, and not a power, unless it was a typo.
     
  10. Jul 8, 2006 #9

    arildno

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    That minus sign was removed by flipping the fraction. :smile:

    EDIT:
    Oh dear, it was MINUS 4 not that the exponent was minus 4 as I thought..:cry:
     
  11. Jul 8, 2006 #10
    Exactly. :)
     
  12. Jul 8, 2006 #11
    Can this be solved analytically?
     
  13. Jul 8, 2006 #12

    arildno

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    Of course.
    It is a linear equation in the variable log(x).
     
  14. Jul 8, 2006 #13
    Not quadratic?
     
  15. Jul 8, 2006 #14

    arildno

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    It might be..:uhh:
     
  16. Jul 8, 2006 #15

    HallsofIvy

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    Assuming this is
    [tex]\left(\frac{log x}{log 4}\right)^2= \frac{log x^5}{log 4}- 4[/tex]
    and not
    [tex]\left(\frac{log x}{log 4}\right)= \left(\frac{log x^5}{log 4}\right)^{- 4}[/tex]
    as I also initially interpreted it, let y= log x. since log x5= 5 log x, this can be written as
    [tex]\frac{1}{(log 4)^2}y^2- \frac{5}{log 4} y+ 4= 0[/tex]
    a quadratic equation for y.
    ("log 4" is, of course, simply a constant.)
     
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