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I Find This Very Hard Can Somebody Check This Out

  1. Apr 24, 2005 #1
    :mad: 1. Given y= (2e^t + 3e^-t ) / (e^t + 2e^-t)

    find inverse of of the equation from above

    i did the working out to get

    y= ([2t-3] / [2-t]) ^ 1/2

    do i have to apply any log laws or anything like that?

    2. cosh x = 4.5 find sinh x exactly
    i applied the formula
    (coshx)^2 - (sinhx)^2 = 1
    therefore

    - (sinhx)^2 = 1 - (coshx)^2
    (sinhx)^2 = -1 + (4.5)^2
    (sinhx)^2 = 19.25
    sinhx = sqrt 19.25

    is that exactly????

    3. express as partial fractions

    (10x^2 + 14x + 3) / (x^3 + 3x^2 - 4)
    i got
    [7/(x+2)] - [5/(x+2)^2] + [3/(x-1)]

    is that right?

    PLEASE REPLY THANKS
     
  2. jcsd
  3. Apr 24, 2005 #2
    sorry made an error in the 1st question meawn to be
    for my solution

    y= 1/2 ln ([2t-3] / [2 - t])

    not this
    y= ([2t-3] / [2-t]) ^ 1/2
     
  4. Apr 24, 2005 #3
    [tex]
    y = \frac{{2e^t + 3e^{ - t} }}{{e^t + 2e^{ - t} }}
    [/tex]

    Interchange y and t(it isn't necessary to do that but I prefer to).

    [tex]
    t = \frac{{2e^y + 3e^{ - y} }}{{e^y + 2e^{ - y} }}
    [/tex]

    [tex]
    t = \frac{{\frac{{2e^{2y} + 3}}{{e^y }}}}{{\frac{{e^{2y} + 2}}{{e^y }}}}
    [/tex]

    [tex]
    t = \frac{{2e^{2y} + 3}}{{e^{2y} + 2}}
    [/tex]

    [tex]
    te^{2y} + 2t = 2e^{2y} + 3
    [/tex]

    [tex]
    \left( {t - 2} \right)e^{2y} = 3 - 2t
    [/tex]

    [tex]
    e^{2y} = \frac{{3 - 2t}}{{\left( {t - 2} \right)}}
    [/tex]

    [tex]
    y = \log _e \left( {\sqrt {\frac{{3 - 2t}}{{t - 2}}} } \right)
    [/tex]

    Edit: Our answers for the first one are equivalent. To convert mine to yours just multiply the numerator and the denominator of the log argument by negative one and get rid of the square root by placing a factor of (1/2) in front of the log.
     
    Last edited: Apr 24, 2005
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