# Calculus help

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!!!!!

HallsofIvy
Homework Helper
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.

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:
ne1??

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.

HallsofIvy