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Math Help - Logs Help

  1. #1
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    Logs Help

    hi all. i've been stuck on this logarithm question for a while now, and i just can't seem to get the answer.

    the question is
    e^(4x^2 - 2) = 3
    (the 4x^2 - 2 is supposed to be the power of e)
    solve for x to 2 decimals.

    anyways, i keep on getting a positive and negative number answer, but it's the incorrect value. i get like +- 0.88 , so i'm just curious as to how i can solve this question.

    thank you very much
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  2. #2
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    Quote Originally Posted by mneox View Post
    hi all. i've been stuck on this logarithm question for a while now, and i just can't seem to get the answer.

    the question is
    e^(4x^2 - 2) = 3
    (the 4x^2 - 2 is supposed to be the power of e)
    solve for x to 2 decimals.

    anyways, i keep on getting a positive and negative number answer, but it's the incorrect value. i get like +- 0.88 , so i'm just curious as to how i can solve this question.

    thank you very much
    your values are correct ... why do you think they are wrong?
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  3. #3
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    Quote Originally Posted by mneox View Post
    hi all. i've been stuck on this logarithm question for a while now, and i just can't seem to get the answer.

    the question is
    e^(4x^2 - 2) = 3
    (the 4x^2 - 2 is supposed to be the power of e)
    solve for x to 2 decimals.

    anyways, i keep on getting a positive and negative number answer, but it's the incorrect value. i get like +- 0.88 , so i'm just curious as to how i can solve this question.

    thank you very much
    e^{4x^2 - 2} = 3

    4x^2 - 2 = \ln{3}

    4x^2 = 2 + \ln{3}

    x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}

    x = \pm\sqrt{\frac{1}{2} + \frac{1}{4}\ln{3}}.


    Therefore

    x = -\sqrt{\frac{1}{2} + \frac{1}{4}\ln{3}} or x = \sqrt{\frac{1}{2} + \frac{1}{4}\ln{3}}.


    Now put these into your calculator for the decimal approximations.
    Last edited by Prove It; November 28th 2009 at 07:39 PM.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    e^{4x^2 - 2} = 3

    4x^2 - 2 = \ln{3}

    4x^2 = 2 + \ln{3}

    x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}

    x = \pm\left(\frac{1}{2} + \frac{1}{4}\ln{3}\right).

    forget a square root?



    Therefore

    x = -\frac{1}{2} - \frac{1}{4}\ln{3} or x = \frac{1}{2} + \frac{1}{4}\ln{3}.


    Now put these into your calculator for the decimal approximations.
    ...
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  5. #5
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    Quote Originally Posted by skeeter View Post
    your values are correct ... why do you think they are wrong?
    for one, we have a answer key and it says that the correct value is +- 1.76.

    but if what i got was correct, maybe it's just the answer key is incorrect?

    Quote Originally Posted by Prove It View Post
    e^{4x^2 - 2} = 3

    4x^2 - 2 = \ln{3}

    4x^2 = 2 + \ln{3}

    x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}

    x = \pm\left(\frac{1}{2} + \frac{1}{4}\ln{3}\right).


    Therefore

    x = -\frac{1}{2} - \frac{1}{4}\ln{3} or x = \frac{1}{2} + \frac{1}{4}\ln{3}.


    Now put these into your calculator for the decimal approximations.
    hi, thanks for the reply.

    i was just wondering how you got the same answer but with a +/- after you square root?

    x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}

    x = \pm\left(\frac{1}{2} + \frac{1}{4}\ln{3}\right)
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  6. #6
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    try your solutions in the original equation , then try the "book" answers.

    see for yourself which are correct.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    ...
    Yes I did - edited.
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