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Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
1. Inverse Functions
x+3
Consider the function f (x) =
x−2
(a) Find the inverse function f −1 (x).
y = f (x) = x + 3x − 2
y(x − 2) = x + 3
yx − 2y = x + 3
yx − x = 2y + 3
x(y − 1) = 2y + 3
x = f −1 (y) =
2y + 3
y−1
2y + 3
2x + 3
or you could say f −1 (x) =
y−1
x−1
(b) To check this is an inverse function notice that
Thus x = f −1 (y) =
2x + 3
+3
2x + 3
x−1
f (f −1 )(x) = f
=
2x + 3
x−1
−2
x−1


2x + 3
+
3
 x−1
 x−1


=
 x−1
2x + 3
−2
x−1
(2x + 3) + 3(x − 1)
=
(2x + 3) − 2(x − 1)
2x + 3 + 3x − 3
5x
=
=
=x
2x + 3 − 2x + 2
5
And also
x+3
2
+3
x+3
x−2
−1
−1
f (f (x)) = f
=
x+3
x−2
−1
x−2


x+3
2
+
3

 x−2
x−2

=
 x+3
 x−2
−1
x−2
2(x + 3) + 3(x − 2)
=
(x + 3) − (x − 2)
2x + 6 + 3x − 6
5x
=
=
=x
x+3−x+2
5
Thus f (f −1 (x)) = x and f −1 (f (x)) = x.
1
Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
x+3
(c) The domain of f (x) =
is x − 2 6= 0 or x 6= 2.
x−2
2x + 3
is x − 1 6= 0 or x 6= 1.
The domain of f −1 (x) =
x−1
Since the range of f (x) is the domain of f −1 (x) and visa versa, we get that
x+3
• f (x) =
has domain
x−2
{x | x 6= 2} = (−∞, 2) ∪ (2, ∞)
2
And has the range
2x + 3
• f −1 (x) =
has domain
x−1
{x
| x 6= 1} = (−∞, 1) ∪ (1, ∞)
{x
| x 6= 1} = (−∞, 1) ∪ (1, ∞)
{x
| x 6= 2} = (−∞, 2) ∪ (2, ∞)
And has the range
2. Logarithms
Calculate out the following logarithms by first rewriting the equation in exponential form:
(a)
y = log6 (216)
6y = 6log6 (216) = 216
6y = 63
y=3
Thus log6 (216) = 3
(b)
y = log5
log5
5y = 5
1
625
!
1
625 = 1
625
1
= 5−4
54
y = −4
5y =
Thus log5 (1/625) = −4
(c)
y = ln(e25 )
ey = 3ln(e
25
= e25
ey = e25
y = 25
Thus ln(e25 ) = 25.
Another method is since ln(x) is the inverse function of ex , we have the inverse function property ln(ex ) = x.
Thus
ln(e25 ) = 25
Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
3. Logarithmic Functions
3
Consider the function f (x) = ln(3 − x) + 5 = ln(−(x − 3)).
(a) The parent function is p(x) = ln(x) and this function is obtained from ln(x) by
• Horizontal reflection about the y-axis.
• Move 3 units to the right.
• Move 5 units up.
(b) The domain is all reals such that 3 − x > 0 or x < 3.
The range is all real numbers (since the parent function ln(x) has all real numbers as the range and shifting all
real numbers up 5 is still all real numbers).
The vertical asymptote occurs when 3 − x = 0 or x = 3.
(c) The graph of f (x) is
4. Properties of Logarithms
(a) Use Properties of logarithms to expand
3x2
ln
= ln(3x2 ) − ln((x + 1)10 ) = ln(3) + ln(x2 ) − 10 ln(x + 1) = ln(3) + 2 ln(x) − 10 ln(x + 1)
(x + 1)10
(b) Use Properties of logarithms to combine and simplify
3 log(x) − log(x2 − 1) + 2 log(x − 1) = log(x3 ) + log((x − 1)2 ) − log(x2 − 1)
3
x (x − 1)2
= log
x2 − 1
3
x (x − 1)2
= log
(x − 1)(x + 1)
3
x (x − 1)
= log
x+1
Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
5. Exponential and Logarithmic Equations
4
Solve the following equations:
(a)
21−x = 52x
ln(21−x ) = ln(52x )
(1 − x) ln(2) = 2x ln(5)
ln(2) − x ln(2) = 2x ln(5)
ln(2) = 2x ln(5) + x ln(2)
ln(2) = x(2 ln(5) + ln(2))
x=
ln(2)
ln(2)
ln(2)
=
=
≈ 0.177
2 ln(5) + ln(2)
ln(52 · 2)
ln(50)
(b)
10
=2
1 + e−x
10 = 2(1 + e−x )
5 = 1 + e−x
4 = e−x
ln(4) = −x
x = − ln(4) ≈ −1.386
(c)
e2x − ex − 12 = 0
(ex )2 − (ex ) − 12 = 0
(ex − 4)(ex + 3) = 0
ex = 4
x = ln(4)
or ex = −3
or
No Solutions
Notice that since the range of ex is (0, ∞), there are no solutions to ex = −3. Thus the only solution is x =
ln(4) ≈ 1.386
(d)
log2 (x + 2) + log2 (x − 1) = 2
log2 ((x + 2)(x − 1)) = 2
(x + 2)(x − 1) = 22 = 4
x2 + x − 2 = 4
x2 + x − 6 = 0
(x + 3)(x − 2) = 0
x = −3
or x = 2
Since x = −3 is not in the domain of log2 (x + 2) + log2 (x − 1) it is an extraneous solution.
Thus the only solution is x = 2.