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Math Help - Incorrect equation, please help me fix it.

  1. #1
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    Incorrect equation, please help me fix it.

    My teacher said this is incorrect because I should be taking the natural log of both sides of the equation. How would I do that?

    y = e^3


    y = e^3


    e = 3SQRT(y)


    e = y^1/3


    e^2 = y^2/3

    Thanks
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  2. #2
    Senior Member furor celtica's Avatar
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    what are you trying to find? if you are looking for y, you have it already. e^3 is a constant.
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  3. #3
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    Quote Originally Posted by bobbyboy1111 View Post
    My teacher said this is incorrect because I should be taking the natural log of both sides of the equation. How would I do that?

    y = e^3


    y = e^3


    e = 3SQRT(y)


    e = y^1/3


    e^2 = y^2/3

    Thanks

    remember that when you have y = b ^ x, this can be written as log_b y = x.

    In this case, b = e, so log_e (y) = 3. Log to base e is written ln, so this becomes ln (y) = 3.

    But, it's not clear to me what problem you are really trying to solve. Are you trying to solve for y? If so, just compute e^3 on your calculator, and that's the answer.
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  4. #4
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    Quote Originally Posted by QM deFuturo View Post
    remember that when you have y = b ^ x, this can be written as log_b y = x.

    In this case, b = e, so log_e (y) = 3. Log to base e is written ln, so this becomes ln (y) = 3.

    But, it's not clear to me what problem you are really trying to solve. Are you trying to solve for y? If so, just compute e^3 on your calculator, and that's the answer.
    My teacher told me to solve by taking the natural log of both sides of the equation, that's it, finito.

    So, would this be correct?:

    y = e^3
    log_e(y) = 3
    ln(y) = 3

    Can this be simplified any further?

    Thanks.
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  5. #5
    Senior Member furor celtica's Avatar
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    no. natural log of e^3 is 3 (obviously)
    natural log of y is lny (obviously)
    there is nothing more to it in this case.
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  6. #6
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    Quote Originally Posted by furor celtica View Post
    no. natural log of e^3 is 3 (obviously)
    natural log of y is lny (obviously)
    there is nothing more to it in this case.
    So would that be:

    y = e^3

    ln(y) = 3?
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  7. #7
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by bobbyboy1111 View Post
    My teacher said this is incorrect because I should be taking the natural log of both sides of the equation. How would I do that?

    y = e^3


    y = e^3


    e = 3SQRT(y)


    e = y^1/3


    e^2 = y^2/3

    Thanks
    y=e^3

    \log _e y = \log _e (e^3) note that log_e = ln

    \ln y = 3 \ln (e)

    \ln y = 3
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  8. #8
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    Quote Originally Posted by bobbyboy1111 View Post
    My teacher said this is incorrect because I should be taking the natural log of both sides of the equation. How would I do that?

    y = e^3


    y = e^3


    e = 3SQRT(y)


    e = y^1/3


    e^2 = y^2/3


    Thanks
    Posted and discussed here: http://www.mathhelpforum.com/math-he...n-correct.html

    Thread closed.
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