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Thread: Exponential turning point q

  1. #1
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    Exponential turning point q

    Find the co ordinates of the turning point of the curve
    $\displaystyle y = e^2^x - 3e^x$
    and determine whether this point is a local max or local min.

    I've differentiated this and put it = to 0 and got
    $\displaystyle 2e^2^x - 3e^x$ is that right? In the answer it says that x is loge^3/2 how do I change the e to log?
    Thanks
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Yes, so far so good.

    Equate to zero now.

    $\displaystyle 2e^{2x} - 3e^x=0$

    If you want, you can make a substitution of the type $\displaystyle y = e^x$ to help you.
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  3. #3
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    $\displaystyle e^x (2e^x-3) = 0$
    $\displaystyle e^x =0 (cant be zero?) 2e^x = 3$
    $\displaystyle log2e^3 = x$
    but in the answer it says$\displaystyle loge^3/2$
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  4. #4
    MHF Contributor Unknown008's Avatar
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    If you mean this:

    $\displaystyle log_e \dfrac32$

    Then yes, it's the right answer.

    $\displaystyle e^x = \dfrac32$

    $\displaystyle log\ e^x = log \dfrac32$

    $\displaystyle xlog\ e = log \dfrac32$

    $\displaystyle x = \dfrac{log\ \dfrac32}{log\ e}$

    $\displaystyle x = log_e\ \dfrac32$
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  5. #5
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    How did you do the last 2 lines? dont get how e ends up on top.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    I thought you were told this law...

    You have:

    $\displaystyle \dfrac{log_a \ b }{log_a\ c} = \log_c\ b$

    If in the fraction, the two logs have the same base, then you can change the base like I just did. This also applies in the reverse way.

    Here are some other ways to change the base:

    $\displaystyle \dfrac{1}{log_a\ b} = log_b\ a$

    If:

    $\displaystyle a^b = c$

    Then;

    $\displaystyle b = log_a\ c$
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