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Simple log problem

  1. Sep 18, 2004 #1

    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.

    Last edited: Sep 18, 2004
  2. jcsd
  3. Sep 18, 2004 #2


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    How about this

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

    Maybe i should arrange

    [tex] (4^x)^2 + 1 = \frac{5}{2}(4^x) [/tex]

    [tex] x= 4^x [/tex]

    [tex] x^2 + 1 = \frac{5}{2}x [/tex]
    Last edited: Sep 18, 2004
  4. Sep 19, 2004 #3
    &log(AB)=log\ A+log\ B\\
    &There\ isn't\ anything\ like\ this:\\
  5. Sep 20, 2004 #4
    [tex] x= 4^x [/tex] should not be used. I see where you are going here. But this is wrong. Instead we should assign it to a different variable like [tex] y= 4^x [/tex]. So,

    [tex] y^2 + 1 = \frac{5}{2}y [/tex]
  6. Sep 20, 2004 #5


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    It's true, i was just reminding him of Ax^2 + Bx + C, but thanks for pointing it out.
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