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Some Annoying Logarithms

  1. Aug 22, 2007 #1
    1. The problem statement, all variables and given/known data

    3^(x - 2) = 30

    (x - 2)^3 = 30

    log (base 4) (x + 3) + log (base 4) (x - 3) = 2

    the square root of: (4^(x + 3) / 16^x) = 32

    log (base 9) x = 1.5

    2. Relevant equations

    NONE

    3. The attempt at a solution

    I haven't done this since last year. Someone refresh my memory. Please show me how to do it and fast fast fast!!! Thanks in advance.
     
  2. jcsd
  3. Aug 22, 2007 #2
    you need to show work b4 help is given.
     
  4. Aug 22, 2007 #3
    i struggled but finally got every one but this one:

    log (base 4) (x + 3) + log (base 4) (x - 3) = 2

    just get me started please. thanks.
     
  5. Aug 22, 2007 #4
    i know there is a rule that if they are the same base logs, u can do something to them but idk really!
     
  6. Aug 22, 2007 #5
    the way logs are read ... log(base)y=x

    the base must be raised to some power x to get y

    show an attempt at all of your questions and i will work them all out!
     
  7. Aug 22, 2007 #6
    ok i think i vaguely remember that if u add logs with the same base, you can multiply what is in the log. that woiuld make it:

    log base 4 of x^2 - 9 = 2

    meaning 16 = x^2 - 9

    meaning x^2 = 25

    meaning x = 5

    hooray, thanks!
     
  8. Aug 22, 2007 #7
    excellent :) the first 2 are ez since now you've solved this one ... the square root one, i'll work

    square both sides

    4^(x+3)/4^(2x) = 32^2

    4^(x+3-2x) = 32^2 (simplify your exponent, take the log of both sides, use one of the rules of the power rule, bring down the exponent)

    x = 3 - log(32^2)/log(4)
     
    Last edited: Aug 22, 2007
  9. Aug 22, 2007 #8
    done em all, thx roco
     
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