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Math Help - Logarithm problem

  1. #1
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    Logarithm problem

    Hi, I've been attempting this problem for a while now, and I can't see how they got the solution.
    Problem: 3^(x-1)=2^(2x-1)
    Solution: (log 3-log 2) / (log 3-2log 2)

    Can anyone show me the method for solving such an equation?
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  2. #2
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    Quote Originally Posted by ima9rd View Post
    Hi, I've been attempting this problem for a while now, and I can't see how they got the solution.
    Problem: 3^(x-1)=2^(2x-1)
    Solution: (log 3-log 2) / (log 3-2log 2)

    Can anyone show me the method for solving such an equation?
    Hello,

    I've attached an image of the calculations (my result differs slightly from your solution, but I assume that you've made a typo ):
    Attached Thumbnails Attached Thumbnails Logarithm problem-exp_glg.gif  
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hello,

    I've attached an image of the calculations (my result differs slightly from your solution, but I assume that you've made a typo ):
    That's a different way of doing it, but I understand it. Thanks . Oh, and your answer was the same. log 4 = log 2^2 = 2log 2
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  4. #4
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    Quote Originally Posted by ima9rd View Post
    That's a different way of doing it, but I understand it. Thanks . Oh, and your answer was the same. log 4 = log 2^2 = 2log 2
    Hi,

    of course you are right... I should have taken a cup of coffee first to get awake before I start answering your questions.

    EB
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  5. #5
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    Quote Originally Posted by ima9rd View Post
    Hi, I've been attempting this problem for a while now, and I can't see how they got the solution.
    Problem: 3^(x-1)=2^(2x-1)
    Solution: (log 3-log 2) / (log 3-2log 2)

    Can anyone show me the method for solving such an equation?
    Quote Originally Posted by ima9rd View Post
    That's a different way of doing it, but I understand it. Thanks . Oh, and your answer was the same. log 4 = log 2^2 = 2log 2
    Another slightly different way of doing it:

    3^(x - 1) = 2^(2x - 1)

    log [3^(x - 1)] = log [2^(2x - 1)]

    (x - 1) log 3 = (2x - 1) log 2

    x log 3 - log 3 = 2x log 2 - log 2

    x log 3 = 2x log 2 - log 2 + log 3

    x log 3 - 2x log 2 = log 3 - log 2

    x(log 3 - 2 log 2) = log 3 - log 2

    x = {log 3 - log 2)/(log 3 - 2 log 2)
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  6. #6
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    Hello, ima9rd!

    I'll solve it the traditional way . . .


    Problem: . 3^(x-1) .= .2^(2x-1)

    Solution: .[log(3) - log(2)] / [(log(3) -2Ělog(2)]

    Take the log of both sides:

    . . . . . . . . . . . log[3^
    (x-1)] .= .log[2^(2x-1)]

    . . . . . . . . . . . (x-1)Ělog(3) .= .(2x-1)Ělog(2)

    . . . . . . . . xĚlog(3) - log(3) .= .2xĚlog(2) - log(2)

    . . . . . . xĚlog(3) - 2xĚlog(2) .= .log(3) - log(2)

    Factor: .x[log(3) - 2Ělog(2)] .= .log(3) - log(2)


    . . . . . . . . . . . . . . log(3) - log(2)
    Therefore: .x . = . -------------------
    . . . . . . . . . . . . . log(3) - 2Ělog(2)

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