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Thread: Solve for x

  1. #1
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    Solve for x

    How to solve for x ?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by totalnewbie
    How to solve for x ?
    You should check the question, as set there is no solution for
    real $\displaystyle x$ (sketch the curve for $\displaystyle x>0$ to see why).

    Problem was: Solve

    $\displaystyle
    \ln(x)=\frac{3x^2}{2 \mathrm{e}}
    $

    RonL
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  3. #3
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    It seems to me that I have lost. My goal was to find common tangent for two functions: f(x)=x^2/2e and g(x)=ln(x)
    It was also said that x-coordinate is sqrt(e^1) for one tangent point.
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  4. #4
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    Quote Originally Posted by totalnewbie
    It seems to me that I have lost. My goal was to find common tangent for two functions: f(x)=x^2/2e and g(x)=ln(x)
    It was also said that x-coordinate is sqrt(e^1) for one tangent point.
    Some trial and error might help here.

    Trial 1
    Assume the $\displaystyle x=\sqrt{e}$ point is the tangency point on $\displaystyle f(x)$
    Then we can find the slope of the tangent at $\displaystyle x=\sqrt{e}$, and
    the corresponding value of $\displaystyle f(x)$, which will allow us to find
    the equation of the tangent. Now check that this line is a tangent
    to $\displaystyle g(x)$, if it is our job is done, if not proceed to Trial 2

    Trial 2
    Assume the $\displaystyle x=\sqrt{e}$ point is the tangency point on $\displaystyle g(x)$,
    then proceed as in Trial 1, but with the roles of $\displaystyle f$ and $\displaystyle g$
    interchanged.

    RonL
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  5. #5
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    I have tried it.
    The green line doesn't touch the red parabola. But it should touch.
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  6. #6
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    Quote Originally Posted by totalnewbie
    It seems to me that I have lost. My goal was to find common tangent for two functions: f(x)=x^2/2e and g(x)=ln(x)
    It was also said that x-coordinate is sqrt(e^1) for one tangent point.
    Hello,

    if there exists a common tangent then the gradient of both functions must be equal. So first calculate the drivative of both functions:
    $\displaystyle \frac{df}{dx}=\frac{1}{e} \cdot x$
    $\displaystyle \frac{dg}{dx}=\frac{1}{x}$

    Both are equal: $\displaystyle \frac{1}{e} \cdot x=\frac{1}{x}$

    Solve for x and you'll get: $\displaystyle x=\sqrt{e}$

    Complete the coordinates:
    $\displaystyle f(\sqrt{e})=\frac{1}{2}$
    $\displaystyle g(\sqrt{e})=\frac{1}{2}$

    That means that the graphs of both functions have one common point: the tangent point. There exists one tangent (that's a special case, normally there must be two tangents!)

    Use the point-slope-formula: $\displaystyle t: y=\frac{1}{\sqrt{e}} \cdot (x-\sqrt{e})+\frac{1}{2}$

    I've attached a drawing to demonstrate my results.

    Greetings

    EB
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  7. #7
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    Quote Originally Posted by totalnewbie
    I have tried it.
    The green line doesn't touch the red parabola. But it should touch.
    Hello,

    you have typed the function f like this: x^2/2e. Then your computer will calculate $\displaystyle \frac{x^2}{2} \cdot e$.

    You have to use parantheses: $\displaystyle \frac{x^2}{(2 \cdot e)}$

    Good luck.

    Greetings

    EB
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by totalnewbie
    I have tried it.
    The green line doesn't touch the red parabola. But it should touch.
    So now switch to the other case (where the tangent at $\displaystyle x=\sqrt{e}$ is to the other curve).

    RonL
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