Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Pyrrhus

    User Avatar
    Homework Helper

    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
    [tex]
    \begin{multline*}
    \begin{split}
    &log(AB)=log\ A+log\ B\\
    &There\ isn't\ anything\ like\ this:\\
    &log(A+B)=log(AB)\\
    &4^x+4^{-x}=4^x+\frac{1}{4^x}=\frac{4^{2x}+1}{4^x}\\
    &4^x*4^{-x}=4^{x-x}=1\\
    \end{split}
    \end{multline*}
    [/tex]
     
  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

    Pyrrhus

    User Avatar
    Homework Helper

    It's true, i was just reminding him of Ax^2 + Bx + C, but thanks for pointing it out.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook