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Math Help - logarithmic question

  1. #1
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    logarithmic question

    Code:
    need some help urgently
    
    Solve the equation.
    (16/9)^{x-1} = (3/4)^{x-1}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by math321 View Post
    need some help urgently

    Solve the equation.
    (16/9)^{x-1} = (3/4)^{x-1}
    log both sides so that you can bring down the powers.

    \displaystyle \log (16/9)^{x - 1} = \log (3/4)^{x - 1}

    \displaystyle \Rightarrow (x - 1) \log (16/9) = (x - 1) \log (3/4)

    now continue (note that \displaystyle \log (16/9) and \displaystyle \log (3/4) are constants). where can you go from there?
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  3. #3
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    So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by math321 View Post
    So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x
    yes, but just leave them as log (16/9) and log (3/4), no need to evaluate them. so you're on the right track. multiplying out and getting all x's on one side would be the way to proceed.
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  5. #5
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    can u verify this anser for me please

    (16/9)^(x-1) = (9/16)^-(x-1) = (9/16)^(1-x) = ((3/4)^2)^(1-x) = (3/4)^(2(1-x))

    (3/4)^(2(1-x)) = (3/4)^(x-1)

    2(1-x) = (x-1)
    2-2x = x-1
    3 = 3x
    x = 1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by math321 View Post
    can u verify this anser for me please

    (16/9)^(x-1) = (9/16)^-(x-1) = (9/16)^(1-x) = ((3/4)^2)^(1-x) = (3/4)^(2(1-x))

    (3/4)^(2(1-x)) = (3/4)^(x-1)

    2(1-x) = (x-1)
    2-2x = x-1
    3 = 3x
    x = 1
    yes, that works!
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