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Math Help - Log problem

  1. #1
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    Log problem

    The solutions to (log_3x)^2-3log_3x-4=0 are x\:=\:,

    No clue how to work with logs...
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  2. #2
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    Quote Originally Posted by davidman View Post
    The solutions to (log_3x)^2-3log_3x-4=0 are x\:=\:,

    No clue how to work with logs...
    Set u = log_3(x)

    u^2-3u-4=0

    Solve the quadratic for u. Remember only positive values of u will give positive values of x
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by davidman View Post
    The solutions to (log_3x)^2-3log_3x-4=0 are x\:=\:,

    No clue how to work with logs...
    To solve this first, take this as a normal quadratic.

    If we replace log_3x with Y, we get:

    Y^2 - 3Y - 4=0

    Solving this for Y gives you Y = (-1,4).

    If we replace Y with log_3x, we get the following two equations to solve:

    log_3x = -1

    and

    log_3x = 4.

    You then need to solve these for x.

    Have you worked with logs at all? For example, log_3 81 = 4, this is the same as 3^4 = 81. Does this ring any bells?
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  4. #4
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    Quote Originally Posted by craig View Post
    Have you worked with logs at all? For example, log_3 81 = 4, this is the same as 3^4 = 81. Does this ring any bells?
    Nope, completely 100% brand new to logs... but I guess that makes sense.

    log_xM=N \: \:, \: \: x^N=M

    thanks!

    log_3x=-1\:\:,\:\:3^{-1}=x\:\:,\:\:x=\frac{1}{3}

    log_3x=4\:\:,\:\:3^4=x\:\:,\:\:x=81
    Last edited by davidman; January 31st 2010 at 01:07 PM.
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by davidman View Post
    Nope, completely 100% brand new to logs... but I guess that makes sense.

    log_xM=N \: \:, \: \: x^N=M

    thanks!

    log_3x=-1\:\:,\:\:3^-1=x\:\:,\:\:x=\frac{1}{3}

    log_3x=4\:\:,\:\:3^4=x\:\:,\:\:x=81
    Yep
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