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Log problems

  1. Sep 28, 2008 #1
    Here's 2 I couldn't get.


    log3x2 = log481/2
    Solve for x

    For the first one I thought I should solve what's in the exponent first and then solve 8 to that power, but I can't figure out what to do with the exponent.

    I'm even more clueless on the second one/ The bases are different, and I can't out what to do with the x2 and 81/2

    Can someone please explain this to me? What properties should I use to solve these problems?

    I'm supposed to do these without a calculator, and that's what I'm trying to do.
  2. jcsd
  3. Sep 28, 2008 #2


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    Hi musicfairy! :smile:
    Hint: 8 = 23, and (ab)c = … ? :smile:
    Hint: log48 = … ? :smile:
  4. Sep 28, 2008 #3

    So I tried to follow the hints you gave me and came up with these. I might have made a few new properties..

    8log27 = 23log27
    So 2 and log2 kill each other and kill each other and I end up with 21.

    For the second one I got

    log481/2 = log4(2/3)

    log4(3/2) / log4 1 = 3/2

    2log3x = 3/2

    lnx/ln3 = 3/4

    x/3 = e3/4
    x = 3e3/4

    This problem is tricky...
  5. Sep 28, 2008 #4


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    Hi musicfairy! :smile:
    Goodness … you musicfairies are violent!

    "kill each other"? :rolleyes:

    Hint: 23log27 = (2log27)3 = … ? :smile:
    Not following any of that :confused:

    For example, log41 = 0, isn't it? :smile:

    Answer my original question: log48 = … ?
  6. Sep 28, 2008 #5
    Well, cancel is a better word. I was only quoting someone from class.

    23log27 would equal 73???

    log48= 3/2
  7. Sep 28, 2008 #6


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    good fairy!

    :biggrin: Woohoo! :biggrin:

    And so log481/2 = … ? :smile:
  8. Sep 28, 2008 #7


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    No, you are correct that the exponential and logarithm with base 2, because they are inverse functions, "kill each other" but you are handling the exponent incorrectly.
    [tex]8^{log_2(7)}= (2^3)^{log_2(7)}= 2^{3log_2(7)}[/tex]
    But that is NOT 3*7. 3 log2(7)= log2(73).
    [tex]2^{3 log_2(7)}= 2^{log_2(7^3)}[/tex]
    Now, what is that?

    Since 8= 23 and 2= 41/2, it is true that 8= 43/2 but htat means 81/2= 4(3/2)(1/2)= 43/4. And now, log4(81/2= log4(43/4)= 3/4.

    ??? log4 1= 0. log to any base of 1= 0 because a0= 1 for any positive number a- and you can't divide by 0. It is certainly NOT true that log(a)/log(b)= a/b !

    Is this a different problem now? Are you thinking that eln a/ln b= a/b? That is definitely not true! It is far easier to use the basic definition of logarithm: If y= loga(x), then x= ay. If log3(x)= 3/4, then x= what?
  9. Sep 28, 2008 #8
    Trying to follow your instructions...

    log3x2 = 3/4
    x2 = 33/4
    x = 33/8

    Is it right now?
  10. Sep 28, 2008 #9


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    Your fairy logfather says … yes! :smile:
  11. Sep 28, 2008 #10
    Thanks for all the help and words of wisdom. Now I'll go review log properties.
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