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Math Help - Solving Equations using Logs

  1. #1
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    Solving Equations using Logs

    Hi guys, I am unsure how this book has come about with the answer to this question as I thought I had it worked out.

    Question:

    Use the substitution y = 2^x to solve the following equation giving your answer in exact form involving logarithms to the base ten.

    2^(2x) - 2^(x+3) + 15 = 0

    My work goes like this:

    y^2 - 2^(x+3) + 15 = 0
    y^2 + 15 = 2^(x+3) <- Expand the log
    log y^2 + log 15 = log 2^(x+3)
    = (x+3)log 2
    = xlog 2 + 3log2
    = log 2^x + log 2^3
    = log y + log 8 <- Rearrange the equation
    log y^2 - log y = log 8 - log 15
    log ((y^2)/y) = (log 8/15)
    y = 8/15

    The answers however state the following and I am unsure how they come about as I thought my working was sound.

    log 3 / log 2 & log 5 / log 2

    Any help would be greatly appreciated.

    Cheers

    Ben
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  2. #2
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    You don't need logarithms.

    \displaystyle 2^{2x} - 2^{x + 3} + 15 = 0

    \displaystyle \left(2^x\right)^2 - 2^3\cdot 2^x + 15 = 0.

    Now make the substitution suggested and solve the resulting quadratic.
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  3. #3
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    Quote Originally Posted by MuNch View Post
    Hi guys, I am unsure how this book has come about with the answer to this question as I thought I had it worked out.

    Question:

    Use the substitution y = 2^x to solve the following equation giving your answer in exact form involving logarithms to the base ten.

    2^(2x) - 2^(x+3) + 15 = 0

    My work goes like this:

    y^2 - 2^(x+3) + 15 = 0
    But you didn't make the substitution completely! 2^{x+ 3}= 2^x2^3= 8(2^x)
    Your equation should be y^2+ 8y+ 15= 0
    which is a simple quadratic equation.

    y^2 + 15 = 2^(x+3) <- Expand the log
    log y^2 + log 15 = log 2^(x+3)
    and this is also wrong: log(a+ b) is NOT equal to log(a)+ log(b).

    = (x+3)log 2
    = xlog 2 + 3log2
    = log 2^x + log 2^3
    = log y + log 8 <- Rearrange the equation
    so NOW you have finished the substitution- but the left side, log(y^2)+ log(15), is still wrong.

    log y^2 - log y = log 8 - log 15
    log ((y^2)/y) = (log 8/15)
    y = 8/15

    The answers however state the following and I am unsure how they come about as I thought my working was sound.

    log 3 / log 2 & log 5 / log 2

    Any help would be greatly appreciated.

    Cheers

    Ben
    Last edited by HallsofIvy; March 16th 2011 at 03:12 AM.
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