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Math Help - Natural Log

  1. #1
    Junior Member BobBali's Avatar
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    Natural Log

    Hi All, please see if my steps are alright below:

    Solve for x

    e^{2x} - 7e^x + 12

    e^{2x} - 7e^x = -12

    e^{2x} - (e^x)^{7} = -12

    ln(\frac{e^{2x}}{e^(7x)} = e^{-12}

    ln(e^{-5x} = e^{-12}

    -5x = -12   \rightarrow x = \frac{12}{5}
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  2. #2
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    Re: Natural Log

    Quote Originally Posted by BobBali View Post
    Hi All, please see if my steps are alright below:

    Solve for x

    e^{2x} - 7e^x + 12

    e^{2x} - 7e^x = -12

    e^{2x} - (e^x)^{7} = -12

    ln(\frac{e^{2x}}{e^(7x)} = e^{-12}

    ln(e^{-5x} = e^{-12}

    -5x = -12   \rightarrow x = \frac{12}{5}
    1. I assume that you mean:

    e^{2x} - 7e^x + 12=0

    If so

    2. Use the substitution y = e^x

    Your equation becomes

    y^2-7y+12=0

    3. Use the quadratic formula to solve for y. Afterwards use the equation of #2 to determine x.
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  3. #3
    MHF Contributor
    Prove It's Avatar
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    Re: Natural Log

    Quote Originally Posted by earboth View Post
    1. I assume that you mean:

    e^{2x} - 7e^x + 12=0

    If so

    2. Use the substitution y = e^x

    Your equation becomes

    y^2-7y+12=0

    3. Use the quadratic formula to solve for y. Afterwards use the equation of #2 to determine x.
    Quadratic formula? Surely it's easier if you note that \displaystyle \begin{align*} y^2 - 7y + 12 &= (y - 3)(y - 4) \end{align*}...
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  4. #4
    Junior Member BobBali's Avatar
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    Re: Natural Log

    Quote Originally Posted by Prove It View Post
    Quadratic formula? Surely it's easier if you note that \displaystyle \begin{align*} y^2 - 7y + 12 &= (y - 3)(y - 4) \end{align*}...
    I get y=4 or 3

    Solving for x using e^x =4. Gives ln4= 1.39 and 1.1

    While my way x= 12/5 = 2.4

    Does this mean i$m wrong??!
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  5. #5
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    Re: Natural Log

    Quote Originally Posted by BobBali View Post
    Does this mean i$m wrong??!
    You have a severe error in the OP.

    It is: 7e^x\ne (e^x)^7.

    And that is only one of several.
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