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Calculus help

  1. Sep 12, 2003 #1
    calculus help plz!!

    if f(x)= (2e^x -8)/(10e^x + 9)
    then wat is f^(-1)(x)?

    i first took the ln of both sides....

    getting lny = ln (2e^x - 8)/(10e^x + 9)

    then using one of the properties i get

    lny = ln (2e^x - 8) - ln (10e^x + 9)

    and from here i get stuck...how do i solve for x?? am i doing it a totally wrong way?? plz help!!!!!
     
  2. jcsd
  3. Sep 12, 2003 #2

    HallsofIvy

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    Pretty much, yeah, you're doing it the wrong way.

    Instead of starting right off with the ln (which doesn't really help, does it?) you might want to simplfy the problem a bit first.

    Since y= (2ex- 8)/(10ex+9),
    (10ex+ 9)y= 2ex- 8
    10exy+ 9= 2ex-8

    Now subtract 9 and 2ex from both sides of the equation:
    10y ex- 2ex= -17 or

    (10y- 2)ex= -17

    Divide both sides by 10y- 2 to isolate the exponential:

    ex= -17/(10y-2)= 17/(2- 10y)

    FINALLY, take the ln of both sides:

    x= ln(17/(2-10y)) so the inverse function is

    f-1(x)= ln(17/(2-10x)) where defined.
     
  4. Sep 12, 2003 #3
    er....wat happened to the y? shouldnt it b 9y when u expand the y thru the brackets??
     
    Last edited: Sep 12, 2003
  5. Sep 12, 2003 #4
    ne1??
     
  6. Sep 13, 2003 #5
    y= (2e^x - 8)/(10e^x + 9)

    (10e^x+ 9)y= 2e^x- 8

    10e^xy + 9y = 2e^x - 8

    10e^xy-2e^x = - (8 + 9y)

    e^x = (9y + 8)/(2 - 10y)

    Take ln on both sides
    x = ln [(9y + 8)/(2 - 10y)]

    so,
    f-1(x) = ln[(9y + 8)/(2 - 10y)]

    Yes, that should be 9y. HallsofIvy only made a careless mistake and the way I do this question is exactly the same as that of HallsofIvy.
     
  7. Sep 14, 2003 #6

    HallsofIvy

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    Well, yeah, I did that to see if you were paying attention.

    You BELIEVE that, don't you?
     
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