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Math Help - Log/Exponential problem

  1. #1
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    Log/Exponential problem

    <br />
\frac{e^x-e^{-x}}{2}=10<br />

    Answer:
    <br />
x = ln (10 + \sqrt{101} )<br />

    What I'm doing (More than likely wrong)

    <br />
\frac{e^x - e}{2^x} = 10<br /> <br />

    <br />
e^x-e=10(2^x)<br />

    <br />
e^x-e=20^x<br />
    Last edited by desiderius1; April 18th 2010 at 10:04 PM.
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  2. #2
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    According to the answer, the given problem is wrong.
    The equation should be equated to 10 rather than 7.
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by desiderius1 View Post
    <br />
\frac{e^x-{\color{blue}e^{-x}}}{{\color{red}2}}=7<br />

    Answer:
    <br />
x = ln (10 + \sqrt{101} )<br />
    This answer is wrong, assuming that you have managed to write down the correct equation that you were asked to solve.

    What I'm doing (More than likely wrong)

    <br />
\frac{e^x - {\color{blue}e}}{{\color{red}2^x}} = 7<br />
    The above equation is very different from the one that you were given.

    <br />
e^x-e=7(2^x)<br />
    <br />
e^x-e=14^x<br />
    That last step is not valid. More generally, you have that in general a\cdot b^n\neq (a\cdot b)^n.
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  4. #4
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    Sorry, there are two identical problems, one equating to 10 and the other 7, I misread while typing.

    The problem is:
    <br />
\frac{e^x-e^{-x}}{2}=10<br />
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by desiderius1 View Post
    Sorry, there are two identical problems, one equating to 10 and the other 7, I misread while typing.

    The problem is:
    <br />
\frac{e^x-e^{-x}}{2}=10<br />
    The easy way to solve this equation would be to recognize that the left side really is \sinh(x), and that the solution is simply x=\arsinh(10).

    Or else you can first multiply the equation by 2e^x to get a quadratic equation for e^x:

    \frac{e^x-e^{-x}}{2}=10 \qquad | \cdot 2e^x

    \left(e^x\right)^2-1=20 e^x \qquad | -20e^x

    \left(e^x\right)^2-20e^x-1=0
    Thus, we get

    \left(e^x\right)_{1,2} = \frac{-(-20)\pm \sqrt{(-20)^2-4\cdot 1\cdot (-1)}}{2\cdot 1}=10\pm \sqrt{101}
    But only the + case is possible, thus we have

    e^x=10+\sqrt{101}\qquad | \ln

    x=\ln(10+\sqrt{101})
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