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Simple log problem

  1. Sep 18, 2004 #1
    --------------------------------------------------------------------------------

    Hello,

    I need help with this problem:


    4^x + 4^-x = 5/2

    My Solution: (Assume we use log with base 4)

    log ( 4^x + 4^-x) = log(5/2)

    = log (4^0) = log (5/2) ????


    I dont see what I am doing wrong. I used the Product Rule. Any help would be greatly appreciated.

    Thanks
     
  2. jcsd
  3. Sep 18, 2004 #2

    jcsd

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    log (4^x + 4^-x) is not equivalent to log(4^x) + log(4^-x).

    try using the fact that 4^x = 4^(2x)*4^-x
     
    Last edited: Sep 18, 2004
  4. Sep 18, 2004 #3

    arildno

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    Dearly Missed

    1.The product rule of logs says:
    log(ab)=log(a)+log(b)
    This is not what you've been using.
    2. You should write, for example:
    [tex]4^{x}=e^{xln(4)},4^{-x}=e^{-xln(4)}[/tex]
    Try to rewrite your original problem with a hyperbolic function..
     
  5. Sep 18, 2004 #4

    Pyrrhus

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    Or simplest

    [tex] (4^x)(4^x) + (4^-^x)(4^x) = \frac{5}{2}(4^x) [/tex]

    Edit: To please arildno :rofl:
     
    Last edited: Sep 18, 2004
  6. Sep 18, 2004 #5

    Tide

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    Try [itex]y = 4^x[/itex].
     
  7. Sep 18, 2004 #6

    arildno

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    Dearly Missed

    I would have said "Or simplest.." :shy: :redface: :cry:
     
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