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Thread: 2 Questions!!

  1. #1
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    2 Questions!!

    questions are on attachment!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Lane
    questions are on attachment!
    1. Solve $\displaystyle 3\ e^{4x-5}=2$

    Divide through by $\displaystyle 3$ to get:

    $\displaystyle e^{4x-5}=2/3$

    Now take natural logs of both sides:

    $\displaystyle
    4x-5=\log_e(2/3)
    $.

    Add $\displaystyle 5$ to both sides:

    $\displaystyle
    4x=\log_e(2/3)+5
    $,

    finally divide through by $\displaystyle 4$:

    $\displaystyle
    x=(\log_e(2/3)+5)/4 = \log_e \left(\root {1/4}\of {2/3} \right) + 5/4
    $

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Lane
    questions are on attachment!
    2. Find $\displaystyle f\circ g$, where $\displaystyle f(x)=4x-5$, and $\displaystyle g(x)=x^2-2x+6$.

    Now

    $\displaystyle
    (f \circ g)(x)=f(g(x))=4 g(x)-5=$$\displaystyle 4(x^2-2x+6)-5=4x^2-8x+19
    $

    RonL
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