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**Unformatted text preview: **nswer anyway, why not use it
earlier to simplify the computations? It is a fair question which we answer unfairly: it’s our book. 6.3 Exponential Equations and Inequalities y=
x= 5ex
ex + 1
5ey
ey + 1 365 Switch x and y x (ey + 1) = 5ey
xey + x = 5ey
x = 5ey − xey
x = ey (5 − x)
x
ey =
5−x
ln (ey ) = ln
y = ln
We claim f −1 (x) = ln x
5−x x
5−x
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
x
of f −1 . We leave this to the reader. To verify our solution graphically, we graph y = f (x) = e5e
x +1
x
and y = g (x) = ln 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. y = f (x) = 5ex
ex +1 and y = g (x) = ln x
5 −x 366 Exponential and Logarithmic Functions 6.3.1 Exercises 1. Solve the following equations analytically.
(a) 3(x−1) = 27
(b) 3(x−1) = 29
(c) 3(x−1) = 2x (h) (l) 2(x 7e2x = (m) (d)
=
x= 1
(e) 8
128
(f) 37x = 814−2x 0.06 12t
12
−5730k = 1
e
2 (j) 1 +
(k) 3 − x) e2x =1 =...

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