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MTH 111 - Spring 2015
§§4.1-4.4 Review Problems (Solutions)
May
4th
Final Exam:
7:00-10:00pm
Name:
x−2
1
+ 3.
2
Determine whether the function is increasing or decreasing on its domain.
Solution:
An exponential function has domain all real numbers. An unaltered exponential function has
range (0, ∞), but the vertical shift of 3 has the effect of “moving” the range up 3 units. Thus
the domain is (−∞, ∞), and the range is (3, ∞).
1. State the domain and range of the exponential function f (x) =
Since the base
1
2
is less than 1, the function is decreasing on its domain.
2. Given the function g(x) = 43−x , find the x-coordinate of the point that has 16 as its ycoordinate. (Hint: Write an equation and use the One-to-One Property.)
Solution:
To find x such that g(x) = 16, we must solve for x in the equation 43−x = 16. However, notice
that
43−x = 16 = (4)2 .
Both sides of the equation now have the same base, so we can employ the One-to-One Property
of exponential functions: the powers must also be the same! Thus 3 − x = 2, so x = 1.
(Indeed, check that g(1) = 16!)
3. Given g(x) = 43−x , find g(5).
Solution:
g(5) = 43−5 = 4−2 =
1
42
=
1
16 ,
so g(5) =
1
.
16
4. Evaluate log4 (64).
Solution:
Here we want to find the power of 4 that yields 64. Since 43 = 64, we conclude the above
logarithm is equal to 3; algebraically, we have that
log4 (64) = x ⇐⇒ 4x = 64 = 43 =⇒ log4 (64) = 3.
1
5. Evaluate log3 ( 27
).
Solution:
1
3 raised to the −3 power yields 27
:
1
1
1
1
x
−3
log3
= x ⇐⇒ 3 =
= 3 =3
=⇒ log3
= −3
27
27
3
27
6. Evaluate ln(1).
Solution:
Recall that loga (1) = 0 for any a > 0, a 6= 1. Equivalently,
ln(1) = x ⇐⇒ ex = 1 = e0 ⇐⇒ ln(1) = 0.
7. Find the domain of the function f (x) = log3 (5x − 7).
Solution:
Since we can only consider the logarithms of strictly positive numbers, we must stipulate that
the interior of this function is always greater than zero. Since 5x − 7 > 0 implies that 5x > 7
or x > 57 , we have that the domain of the function is ( 75 , ∞).
8. Given f (x) = 13 · 5x−4 , find the inverse f −1 .
Solution:
Write y = 31 · 5x−4 . First, switch x and y:
x=
1 y−4
·5 .
3
Multiply both sides by 3:
3x = 5y−4 .
By the definition of a logarithm, we can convert this equation to one of logarithmic form:
log5 (3x) = y − 4.
Finally, add 4 to both sides to yield the following:
y = log5 (3x) + 4.
Thus f −1 (x) = log5 (3x) + 4.
9. Solve log7 (x) = 2.
Solution:
Convert this equation to one of exponential form: log7 (x) = 2 ⇐⇒ x = 72 = 49, so x = 49.
10. Solve ln(x2 ) = ln(2x).
Solution:
Recall the following important property of logarithms:
One-to-One Property of Logarithms For a > 0 and a 6= 1, if loga (x) = loga (y) then x = y.
Since ln(x2 ) = ln(2x), we may conclude x2 = 2x. But then
x2 = 2x ⇐⇒ x2 − 2x = 0 ⇐⇒ x(x − 2) = 0 =⇒ x = 0 or x = 2.
However, we cannot plug x = 0 into the original logarithmic equation since we cannot find the
logarithm of 0. Thus this is an extraneous solution, and our only true solution to the original
equation is x = 2.
11. Rewrite the expression log6 (3) + 14 log6 (x) − 2 log6 (y) − 3 log6 (z) as a single logarithm.
Solution:
First use the Power Rule to bring “coefficients” of logarithms to powers of the interiors of the
functions. Also, factor out −1 from the last two logarithms:
log6 (3) +
1
1
log6 (x) − 2 log6 (y) − 3 log6 (z) = log6 (3) + log6 (x 4 ) − log6 (y 2 ) + log6 (z 3 ) .
4
Using the Product Property, we have that
1
1
log6 (3) + log6 (x 4 ) − log6 (y 2 ) + log6 (z 3 ) = log6 (3x 4 ) − log6 (y 2 z 3 ),
but this difference can be rewritten with the Quotient Property:
1
4
2 3
log6 (3x ) − log6 (y z ) = log6
1 3x 4
.
y2z3
√
73a
12. Rewrite the expression ln
as a sum or difference of multiples of logarithms.
b2
Solution:
Rewrite the quotient with a difference of logarithms. Then apply the Product and Power Rules.
√
√
1
1
73a
= ln(7 3 a) − ln(b2 ) = ln(7) + ln(a 3 ) − ln(b2 ) = ln(7) + ln(a) − 2 ln(b)
ln
2
b
3
13. Solve log7 (x2 ) = log7 (6 − x).
Solution:
Apply the One-to-One Property to the above equation to conclude that x2 = 6 − x. Rewrite
this equation and factor!
x2 = 6 − x ⇐⇒ x2 + x − 6 = 0 ⇐⇒ (x + 3)(x − 2) = 0 =⇒ x = −3, or x = 2.
Although one of our solutions is negative, we will show that neither solution is extraneous.
Plug each potential solution into the original equation to see if the equality is indeed satisfied.
?
X
1) x = 2 =⇒ log7 (22 ) = log7 (6 − 2) ⇐⇒ log7 (4) = log7 (4)
?
X
2) x = −3 =⇒ log7 ((−3)2 ) = log7 (6 − (−3)) ⇐⇒ log7 (9) = log7 (9)
Since each equation is properly satisfied, are solutions are indeed x = −3 and x = 2.
Good luck on the final exam! ,