Solving f(x) for x: Calculus Help Needed!

[sonya]

calculus help please!

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?? please help!

 

[HallsofIvy]

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

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

(10y- 2)e^x= -17

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

e^x= -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.

 

[sonya]

Originally posted by HallsofIvy
Since y= (2e^x- 8)/(10e^x+9),
(10e^x+ 9)y= 2e^x- 8
10e^xy+ 9= 2e^x-8

er...wat happened to the y? shouldn't it b 9y when u expand the y thru the brackets??

 

[sonya]

ne1??

 

[KLscilevothma]

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? shouldn't 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.

 

[HallsofIvy]

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?