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Math Help - Solve Logarithms symbolically

  1. #1
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    Cool Solve Logarithms symbolically

    Ok, this is my first post so here goes.

    I am trying to solve
    Use common or natural logarithms to solve symbolically.

    2e^2x+2= 6
    first I divide both sides by 2
    e^2x+2= 6
    use natural logarithms
    ln e^2x+2 = ln 6
    Simplify
    2x+2 = ln 6
    subtract 2 from both sides
    2x = ln 6-4
    Divide by 2
    x = ln 6-4/2

    I enter this into my Ti-83plus
    ln (6-4)/2
    my answer is .34657 It is wrong I figure it is the way I have entered it into the calc. Anybody have any thoughts of what is wrong?
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  2. #2
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    Quote Originally Posted by AndyHirte View Post
    Ok, this is my first post so here goes.

    I am trying to solve
    Use common or natural logarithms to solve symbolically.

    2e^2x+2= 6
    first I divide both sides by 2
    e^2x+2= 6 Mr F says: This line is wrong. You have to divide all terms by 2, not just the first ....... e^2x + 1 = 3. Personally, my first step would be to subtract 2 from both sides: 2 e^2x = 4 .......

    use natural logarithms
    ln e^2x+2 = ln 6
    Simplify
    2x+2 = ln 6
    subtract 2 from both sides
    2x = ln 6-4
    Divide by 2
    x = ln 6-4/2

    I enter this into my Ti-83plus
    ln (6-4)/2
    my answer is .34657 It is wrong I figure it is the way I have entered it into the calc. Anybody have any thoughts of what is wrong?
    ..
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  3. #3
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    Cool symbolically answer exponential equation

    Thank you for the pointers, still not getting the right answer

    e^2x+1 = 3
    ln e^2x+1 = ln 3
    2x+1 = ln 3
    2x = ln 3-1
    x = ln 3-1/2
    .34657
    still not the right answer. I am doing something else wrong?
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  4. #4
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    Quote Originally Posted by AndyHirte View Post
    Thank you for the pointers, still not getting the right answer

    e^2x+1 = 3
    ln e^2x+1 = ln 3 Mr F says: No. That's totally invalid. Subtract 1 from both sides: e^2x = 2. Now take the log of both sides.

    2x+1 = ln 3
    2x = ln 3-1
    x = ln 3-1/2
    .34657
    still not the right answer. I am doing something else wrong?
    ..
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