Calculus help

  • Thread starter sonya
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  • #1
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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!!!!!
 

Answers and Replies

  • #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.
 
  • #3
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Originally posted by HallsofIvy
Since y= (2ex- 8)/(10ex+9),
(10ex+ 9)y= 2ex- 8
10exy+ 9= 2ex-8
er....wat happened to the y? shouldnt it b 9y when u expand the y thru the brackets??
 
Last edited:
  • #4
23
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ne1??
 
  • #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)]

er....wat happened to the y? shouldnt it b 9y when u expand the y thru the brackets??
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.
 
  • #6
HallsofIvy
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Yes, that should be 9y. HallsofIvy only made a careless mistake
Well, yeah, I did that to see if you were paying attention.

You BELIEVE that, don't you?
 

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