College Algebra, 2013a

6.3 Exponential Equations and Inequalities 455
y =
5e x
e x +1
x =
5e y
e y +1
x (e y +1) = 5e y
xe y + x = 5e y
x = 5e y − xe y
Switch x and y
x = e y (5 − x)
e y x
=
5 − x
( ) x
ln (e y ) = ln
5 − x
( ) x
y = ln
5 − x
)
We claim f −1 (x) =ln(
x
5−x
. To verify this analytically, we would need to verify the compositions
(
f −1 ◦ f ) (x) =x for all x in the domain of f and that ( f ◦ f −1) (x) =x for all x in the domain
of f −1 . We leave this to the reader. To verify our solution graphically, we graph y = f(x) = 5ex
)
and y = g(x) =ln(
x
5−x
on the same set of axes and observe the symmetry about the line y = x.
Note the domain of f is the range of g and vice-versa.
e x +1
( )
y = f(x) = 5ex
e x +1 and y = g(x) =ln x
5−x