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shamil98

  • one year ago

Solve algebraically. \[\frac{ e^x + e^{-x} }{ e^x - e^{-x} } = 5\] I started out by multiply both sides by the bottom fraction and whatnot and took the natural logs of both sides and resulted in error.. haven't done math in months..

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  1. ganeshie8
    • one year ago
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    let \(e^x=u\), rearrange the equation and get a quadratic

  2. triciaal
    • one year ago
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    |dw:1441511606981:dw|

  3. shamil98
    • one year ago
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    Yeah that's what i did originally ^

  4. shamil98
    • one year ago
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    @ganeshie8 i'll try that right now

  5. dan815
    • one year ago
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    ya thats good enuff too

  6. dan815
    • one year ago
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    |dw:1441511754857:dw|

  7. dan815
    • one year ago
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    |dw:1441511833267:dw|

  8. shamil98
    • one year ago
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    OH

  9. shamil98
    • one year ago
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    IM DUMB

  10. triciaal
    • one year ago
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    |dw:1441511696871:dw|

  11. shamil98
    • one year ago
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    thanks guys forgot that e^-x = 1/e^x xD

  12. ganeshie8
    • one year ago
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    if you're not really interested in the numeric value, you could save all that algebra by simply saying \(\coth x = 5 \implies x = \coth^{-1}(5)\)

  13. ganeshie8
    • one year ago
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    http://mathworld.wolfram.com/HyperbolicCotangent.html

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