6.2 Properties of Logarithms

6.2 Properties of Logarithms
6.2 Properties of Logarithms
6.2
437
Properties of Logarithms
In Section 6.1, we introduced the logarithmic functions as inverses of exponential functions and
discussed a few of their functional properties from that perspective. In this section, we explore
the algebraic properties of logarithms. Historically, these have played a huge role in the scientific
development of our society since, among other things, they were used to develop analog computing
devices called slide rules which enabled scientists and engineers to perform accurate calculations
leading to such things as space travel and the moon landing. As we shall see shortly, logs inherit
analogs of all of the properties of exponents you learned in Elementary and Intermediate Algebra.
We first extract two properties from Theorem 6.2 to remind us of the definition of a logarithm as
the inverse of an exponential function.
Theorem 6.3. (Inverse Properties of Exponential and Log Functions) Let b > 0, b 6= 1.
• ba = c if and only if logb (c) = a
• logb (bx ) = x for all x and blogb (x) = x for all x > 0
Next, we spell out what it means for exponential and logarithmic functions to be one-to-one.
Theorem 6.4. (One-to-one Properties of Exponential and Log Functions) Let f (x) =
bx and g(x) = logb (x) where b > 0, b 6= 1. Then f and g are one-to-one. In other words:
• bu = bw if and only if u = w for all real numbers u and w.
• logb (u) = logb (w) if and only if u = w for all real numbers u > 0, w > 0.
We now state the algebraic properties of exponential functions which will serve as a basis for the
properties of logarithms. While these properties may look identical to the ones you learned in
Elementary and Intermediate Algebra, they apply to real number exponents, not just rational
exponents. Note that in the theorem that follows, we are interested in the properties of exponential
functions, so the base b is restricted to b > 0, b 6= 1. An added benefit of this restriction is that it
3/2
eliminates the pathologies discussed in Section 5.3 when, for example, we simplified x2/3
and
obtained |x| instead of what we had expected from the arithmetic in the exponents, x1 = x.
Theorem 6.5. (Algebraic Properties of Exponential Functions) Let f (x) = bx be an
exponential function (b > 0, b 6= 1) and let u and w be real numbers.
• Product Rule: f (u + w) = f (u)f (w). In other words, bu+w = bu bw
• Quotient Rule: f (u − w) =
f (u)
bu
. In other words, bu−w = w
f (w)
b
• Power Rule: (f (u))w = f (uw). In other words, (bu )w = buw
While the properties listed in Theorem 6.5 are certainly believable based on similar properties of
integer and rational exponents, the full proofs require Calculus. To each of these properties of
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Exponential and Logarithmic Functions
exponential functions corresponds an analogous property of logarithmic functions. We list these
below in our next theorem.
Theorem 6.6. (Algebraic Properties of Logarithm Functions) Let g(x) = logb (x) be a
logarithmic function (b > 0, b 6= 1) and let u > 0 and w > 0 be real numbers.
• Product Rule: g(uw) = g(u) + g(w). In other words, logb (uw) = logb (u) + logb (w)
u
u
= g(u) − g(w). In other words, logb
= logb (u) − logb (w)
• Quotient Rule: g
w
w
• Power Rule: g (uw ) = wg(u). In other words, logb (uw ) = w logb (u)
There are a couple of different ways to understand why Theorem 6.6 is true. Consider the product
rule: logb (uw) = logb (u) + logb (w). Let a = logb (uw), c = logb (u), and d = logb (w). Then, by
definition, ba = uw, bc = u and bd = w. Hence, ba = uw = bc bd = bc+d , so that ba = bc+d . By
the one-to-one property of bx , we have a = c + d. In other words, logb (uw) = logb (u) + logb (w).
The remaining properties are proved similarly. From a purely functional approach, we can see
the properties in Theorem 6.6 as an example of how inverse functions interchange the roles of
inputs in outputs. For instance, the Product Rule for exponential functions given in Theorem 6.5,
f (u + w) = f (u)f (w), says that adding inputs results in multiplying outputs. Hence, whatever f −1
is, it must take the products of outputs from f and return them to the sum of their respective inputs.
Since the outputs from f are the inputs to f −1 and vice-versa, we have that that f −1 must take
products of its inputs to the sum of their respective outputs. This is precisely what the Product Rule
for Logarithmic functions states in Theorem 6.6: g(uw) = g(u) + g(w). The reader is encouraged to
view the remaining properties listed in Theorem 6.6 similarly. The following examples help build
familiarity with these properties. In our first example, we are asked to ‘expand’ the logarithms.
This means that we read the properties in Theorem 6.6 from left to right and rewrite products
inside the log as sums outside the log, quotients inside the log as differences outside the log, and
powers inside the log as factors outside the log.1
Example 6.2.1. Expand the following using the properties of logarithms and simplify. Assume
when necessary that all quantities represent positive real numbers.
2
8
3
2. log0.1 10x2
1. log2
3. ln
x
ex
s
100x2
4. log 3
5. log117 x2 − 4
5
yz
Solution.
1. To expand log2
1
8
x
, we use the Quotient Rule identifying u = 8 and w = x and simplify.
Interestingly enough, it is the exact opposite process (which we will practice later) that is most useful in Algebra,
the utility of expanding logarithms becomes apparent in Calculus.
6.2 Properties of Logarithms
439
8
log2
= log2 (8) − log2 (x) Quotient Rule
x
Since 23 = 8
= 3 − log2 (x)
= − log2 (x) + 3
2. In the expression log0.1 10x2 , we have a power (the x2 ) and a product. In order to use the
Product Rule, the entire quantity inside the logarithm must be raised to the same exponent.
Since the exponent 2 applies only to the x, we first apply the Product Rule with u = 10 and
w = x2 . Once we get the x2 by itself inside the log, we may apply the Power Rule with u = x
and w = 2 and simplify.
log0.1 10x2
= log0.1 (10) + log0.1 x2
Product Rule
= log0.1 (10) + 2 log0.1 (x)
Power Rule
= −1 + 2 log0.1 (x)
Since (0.1)−1 = 10
= 2 log0.1 (x) − 1
3 2
3. We have a power, quotient and product occurring in ln ex
. Since the exponent 2 applies
3
to the entire quantity inside the logarithm, we begin with the Power Rule with u = ex
and
w = 2. Next, we see the Quotient Rule is applicable, with
u = 3 and w = ex, so we replace
3
3
ln ex
with the quantity ln(3) − ln(ex). Since ln ex
is being multiplied by 2, the entire
quantity ln(3) − ln(ex) is multiplied by 2. Finally, we apply the Product Rule with u = e and
w = x, and replace ln(ex) with the quantity ln(e) + ln(x), and simplify, keeping in mind that
the natural log is log base e.
ln
3
ex
2
= 2 ln
3
ex
= 2 [ln(3) − ln(ex)]
Power Rule
Quotient Rule
= 2 ln(3) − 2 ln(ex)
= 2 ln(3) − 2 [ln(e) + ln(x)]
Product Rule
= 2 ln(3) − 2 ln(e) − 2 ln(x)
= 2 ln(3) − 2 − 2 ln(x)
Since e1 = e
= −2 ln(x) + 2 ln(3) − 2
4. In Theorem 6.6, there is no mention of how to deal with radicals. However, thinking back to
Definition 5.5, we can rewrite the cube root as a 13 exponent. We begin by using the Power
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Exponential and Logarithmic Functions
Rule2 , and we keep in mind that the common log is log base 10.
s
1/3
100x2
100x2
log 3
=
log
yz 5
yz 5
100x2
1
= 3 log
yz 5
= 13 log 100x2 − log yz 5
= 31 log 100x2 − 31 log yz 5
= 13 log(100) + log x2 − 31 log(y) + log z 5
= 31 log(100) + 31 log x2 − 31 log(y) − 13 log z 5
=
=
=
1
3
2
3
2
3
log(100) + 32 log(x) − 13 log(y) − 53 log(z)
+ 23 log(x) − 31 log(y) − 53 log(z)
log(x) − 13 log(y) − 53 log(z) +
Power Rule
Quotient Rule
Product Rule
Power Rule
Since 102 = 100
2
3
5. At first it seems as if we have no means of simplifying log117 x2 − 4 , since none of the
properties of logs addresses the issue of expanding a difference inside the logarithm. However,
we may factor x2 − 4 = (x + 2)(x − 2) thereby introducing a product which gives us license
to use the Product Rule.
log117 x2 − 4
= log117 [(x + 2)(x − 2)]
Factor
= log117 (x + 2) + log117 (x − 2) Product Rule
A couple of remarks about Example 6.2.1 are in order. First, while not explicitly stated in the above
example, a general rule of thumb to determine which log property to apply first to a complicated
problem is ‘reverse order of operations.’ For example, if we were to substitute a number for x into
the expression log0.1 10x2 , we would first square the x, then multiply by 10. The last step is the
multiplication, which tells us the first log property to apply is the Product Rule. In a multi-step
problem, this rule can give the required guidance on which log property to apply at each step.
The reader is encouraged to look through the solutions to Example 6.2.1 to see this rule in action.
Second, while we were instructed to assume when necessary that all quantities represented positive
real numbers, the authors would be committing
a sin of omission if we failed to point out that, for
instance, the functions f (x) = log117 x2 − 4 and g(x) = log117 (x + 2) + log117 (x − 2) have different
domains, and, hence, are different functions. We leave it to the reader to verify the domain of f
is (−∞, −2) ∪ (2, ∞) whereas the domain of g is (2, ∞). In general, when using log properties to
2
At this point in the text, the reader is encouraged to carefully read through each step and think of which quantity
is playing the role of u and which is playing the role of w as we apply each property.
6.2 Properties of Logarithms
441
expand a logarithm, we may very well be restricting the domain as we do so. One last comment
before we move to reassembling logs from their various bits and pieces. The authors are well aware
of the propensity for some students
to become overexcited and invent their own properties of logs
2
2
like log117 x − 4 = log117 x − log117 (4), which simply isn’t true, in general. The unwritten3
property of logarithms is that if it isn’t written in a textbook, it probably isn’t true.
Example 6.2.2. Use the properties of logarithms to write the following as a single logarithm.
1. log3 (x − 1) − log3 (x + 1)
2. log(x) + 2 log(y) − log(z)
3. 4 log2 (x) + 3
4. − ln(x) −
1
2
Solution. Whereas in Example 6.2.1 we read the properties in Theorem 6.6 from left to right to
expand logarithms, in this example we read them from right to left.
1. The difference of logarithms requires the Quotient Rule: log3 (x−1)−log3 (x+1) = log3
x−1
x+1
.
2. In the expression, log(x) + 2 log(y) − log(z), we have both a sum and difference of logarithms.
However, before we use the product rule to combine log(x) + 2 log(y), we note that we need
to somehow deal with the coefficient 2 on log(y). This can be handled using the Power Rule.
We can then apply the Product and Quotient Rules as we move from left to right. Putting it
all together, we have
log(x) + 2 log(y) − log(z) = log(x) + log y 2 − log(z)
= log xy 2 − log(z)
2
xy
= log
z
Power Rule
Product Rule
Quotient Rule
3. We can certainly get started rewriting 4 log2 (x) + 3 by applying the Power Rule to 4 log2 (x)
to obtain log2 x4 , but in order to use the Product Rule to handle the addition,
we need to
rewrite 3 as a logarithm base 2. From Theorem 6.3, we know 3 = log2 23 , so we get
4 log2 (x) + 3 = log2 x4 + 3
Power Rule
4
3
= log2 x + log2 2
Since 3 = log2 23
= log2 x4 + log2 (8)
= log2 8x4
Product Rule
3
The authors relish the irony involved in writing what follows.
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Exponential and Logarithmic Functions
4. To get started with − ln(x) − 12 , we rewrite
− ln(x) as (−1) ln(x). We can then use the Power
Rule to obtain (−1) ln(x) = ln x−1 . In order to use the Quotient Rule, we need to write 21
√
as a natural logarithm. Theorem 6.3 gives us 21 = ln e1/2 = ln ( e). We have
− ln(x) −
1
2
= (−1) ln(x) − 12
= ln x−1 − 21
Power Rule
= ln x−1 − ln e1/2 Since 12 = ln e1/2
√
= ln x−1 − ln ( e)
−1 x
Quotient Rule
= ln √
e
1
√
= ln
x e
As we would expect, the rule of thumb for re-assembling logarithms is the opposite of what it
was for dismantling them. That is, if we are interested in rewriting an expression as a single
logarithm, we apply log properties following the usual order of operations: deal with multiples of
logs first with the Power Rule, then deal with addition and subtraction using the Product and
Quotient Rules, respectively. Additionally, we find that using log properties in this fashion can
increase the domain of the expression. For example, we leave it to the reader
to verify the domain
x−1
of f (x) = log3 (x−1)−log3 (x+1) is (1, ∞) but the domain of g(x) = log3 x+1 is (−∞, −1)∪(1, ∞).
We will need to keep this in mind when we solve equations involving logarithms in Section 6.4 - it
is precisely for this reason we will have to check for extraneous solutions.
The two logarithm buttons commonly found on calculators are the ‘LOG’ and ‘LN’ buttons which
correspond to the common and natural logs, respectively. Suppose we wanted an approximation to
log2 (7). The answer should be a little less than 3, (Can you explain why?) but how do we coerce
the calculator into telling us a more accurate answer? We need the following theorem.
Theorem 6.7. (Change of Base Formulas) Let a, b > 0, a, b 6= 1.
• ax = bx logb (a) for all real numbers x.
• loga (x) =
logb (x)
for all real numbers x > 0.
logb (a)
The proofs of the Change of Base formulas are a result of the other properties studied in this
section. If we start with bx logb (a) and use the Power Rule in the exponent to rewrite x logb (a) as
logb (ax ) and then apply one of the Inverse Properties in Theorem 6.3, we get
x)
bx logb (a) = blogb (a
= ax ,
6.2 Properties of Logarithms
443
as required. To verify the logarithmic form of the property, we also use the Power Rule and an
Inverse Property. We note that
loga (x) · logb (a) = logb aloga (x) = logb (x),
and we get the result by dividing through by logb (a). Of course, the authors can’t help but point
out the inverse relationship between these two change of base formulas. To change the base of
an exponential expression, we multiply the input by the factor logb (a). To change the base of a
logarithmic expression, we divide the output by the factor logb (a). While, in the grand scheme
of things, both change of base formulas are really saying the same thing, the logarithmic form is
the one usually encountered in Algebra while the exponential form isn’t usually introduced until
Calculus.4 What Theorem 6.7 really tells us is that all exponential and logarithmic functions are
just scalings of one another. Not only does this explain why their graphs have similar shapes, but
it also tells us that we could do all of mathematics with a single base - be it 10, e, 42, or 117. Your
Calculus teacher will have more to say about this when the time comes.
Example 6.2.3. Use an appropriate change of base formula to convert the following expressions
to ones with the indicated base. Verify your answers using a calculator, as appropriate.
1. 32 to base 10
2. 2x to base e
3. log4 (5) to base e
4. ln(x) to base 10
Solution.
1. We apply the Change of Base formula with a = 3 and b = 10 to obtain 32 = 102 log(3) . Typing
the latter in the calculator produces an answer of 9 as required.
2. Here, a = 2 and b = e so we have 2x = ex ln(2) . To verify this on our calculator, we can graph
f (x) = 2x and g(x) = ex ln(2) . Their graphs are indistinguishable which provides evidence
that they are the same function.
y = f (x) = 2x and y = g(x) = ex ln(2)
4
The authors feel so strongly about showing students that every property of logarithms comes from and corresponds
to a property of exponents that we have broken tradition with the vast majority of other authors in this field. This
isn’t the first time this happened, and it certainly won’t be the last.
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Exponential and Logarithmic Functions
3. Applying the change of base with a = 4 and b = e leads us to write log4 (5) =
ln(5)
ln(4) .
Evaluating
ln(5)
ln(4)
this in the calculator gives
≈ 1.16. How do we check this really is the value of log4 (5)?
By definition, log4 (5) is the exponent we put on 4 to get 5. The calculator confirms this.5
4. We write ln(x) = loge (x) = log(x)
log(e) . We graph both f (x) = ln(x) and g(x) =
both graphs appear to be identical.
log(x)
log(e)
y = f (x) = ln(x) and y = g(x) =
5
Which means if it is lying to us about the first answer it gave us, at least it is being consistent.
and find
log(x)
log(e)
6.2 Properties of Logarithms
6.2.1
445
Exercises
In Exercises 1 - 15, expand the given logarithm and simplify. Assume when necessary that all
quantities represent positive real numbers.
1.
ln(x3 y 2 )
2. log2
128
2
x +4
3. log5
√ z
xy
4. log(1.23 × 1037 )
5. ln
7. log√2 4x3
8. log 1 (9x(y 3 − 8))
10. log3
13. log
x2
81y 4
6. log5 x2 − 25
9. log 1000x3 y 5
3
r
√ 100x y
√
3
10
11. ln
4
14. log 1
2
xy
ez
z 3
25
12. log6
!
√
3
4 x2
√
y z
216
x3 y
4
√
3
x
15. ln
√
10 yz
In Exercises 16 - 29, use the properties of logarithms to write the expression as a single logarithm.
16. 4 ln(x) + 2 ln(y)
17. log2 (x) + log2 (y) − log2 (z)
18. log3 (x) − 2 log3 (y)
19.
20. 2 ln(x) − 3 ln(y) − 4 ln(z)
21. log(x) − 13 log(z) + 21 log(y)
22. − 31 ln(x) − 31 ln(y) + 13 ln(z)
23. log5 (x) − 3
24. 3 − log(x)
25. log7 (x) + log7 (x − 3) − 2
26. ln(x) +
1
2
28. log2 (x) + log4 (x − 1)
1
2
log3 (x) − 2 log3 (y) − log3 (z)
27. log2 (x) + log4 (x)
29. log2 (x) + log 1 (x − 1)
2
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Exponential and Logarithmic Functions
In Exercises 30 - 33, use the appropriate change of base formula to convert the given expression to
an expression with the indicated base.
30. 7x−1 to base e
x
2
32.
to base e
3
31. log3 (x + 2) to base 10
33. log(x2 + 1) to base e
In Exercises 34 - 39, use the appropriate change of base formula to approximate the logarithm.
34. log3 (12)
1
37. log4
10
35. log5 (80)
36. log6 (72)
38. log 3 (1000)
39. log 2 (50)
5
3
40. Compare and contrast the graphs of y = ln(x2 ) and y = 2 ln(x).
41. Prove the Quotient Rule and Power Rule for Logarithms.
42. Give numerical examples to show that, in general,
(a) logb (x + y) 6= logb (x) + logb (y)
(b) logb (x − y) 6= logb (x) − logb (y)
logb (x)
x
(c) logb
6=
y
logb (y)
43. The Henderson-Hasselbalch Equation: Suppose HA represents a weak acid. Then we have a
reversible chemical reaction
HA H + + A− .
The acid disassociation constant, Ka , is given by
Kα =
[H + ][A− ]
[A− ]
= [H + ]
,
[HA]
[HA]
where the square brackets denote the concentrations just as they did in Exercise 77 in Section
6.1. The symbol pKa is defined similarly to pH in that pKa = − log(Ka ). Using the definition
of pH from Exercise 77 and the properties of logarithms, derive the Henderson-Hasselbalch
Equation which states
[A− ]
pH = pKa + log
[HA]
44. Research the history of logarithms including the origin of the word ‘logarithm’ itself. Why is
the abbreviation of natural log ‘ln’ and not ‘nl’ ?
45. There is a scene in the movie ‘Apollo 13’ in which several people at Mission Control use slide
rules to verify a computation. Was that scene accurate? Look for other pop culture references
to logarithms and slide rules.
6.2 Properties of Logarithms
6.2.2
447
Answers
1. 3 ln(x) + 2 ln(y)
2. 7 − log2 (x2 + 4)
3. 3 log5 (z) − 6
4. log(1.23) + 37
5.
1
2
ln(z) − ln(x) − ln(y)
6. log5 (x − 5) + log5 (x + 5)
7. 3 log√2 (x) + 4
8. −2 + log 31 (x) + log 31 (y − 2) + log 13 (y 2 + 2y + 4)
9. 3 + 3 log(x) + 5 log(y)
11.
1
4
ln(x) + 14 ln(y) −
13.
5
3
+ log(x) + 12 log(y)
14. −2 + 32 log 1 (x) − log 1 (y) − 12 log 1 (z)
15.
1
3
ln(x) − ln(10) − 12 ln(y) − 12 ln(z)
16. ln(x4 y 2 )
− 14 ln(z)
xy z
17. log2
20. ln
1
4
10. 2 log3 (x) − 4 − 4 log3 (y)
x2
y3 z4
23. log5
21. log
24. log
√
26. ln (x e)
x
29. log2 x−1
32.
2 x
3
2
18. log3
x
125
12. 12 − 12 log6 (x) − 4 log6 (y)
x
y2
22. ln
1000
x
25. log7
q
3
z
xy
x(x−3)
49
31. log3 (x + 2) =
33. log(x2 + 1) =
ln(x2 +1)
ln(10)
34. log3 (12) ≈ 2.26186
35. log5 (80) ≈ 2.72271
36. log6 (72) ≈ 2.38685
37. log4
38. log 3 (1000) ≈ −13.52273
39. log 2 (50) ≈ −9.64824
5
x
y2 z
√
28. log2 x x − 1
30. 7x−1 = e(x−1) ln(7)
2
2
√ 19. log3
√ x y
√
3z
27. log2 x3/2
= ex ln( 3 )
2
3
1
10
≈ −1.66096
log(x+2)
log(3)
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Exponential and Logarithmic Functions
6.3
Exponential Equations and Inequalities
In this section we will develop techniques for solving equations involving exponential functions.
Suppose, for instance, we wanted to solve the equation 2x = 128. After a moment’s calculation, we
find 128 = 27 , so we have 2x = 27 . The one-to-one property of exponential functions, detailed in
Theorem 6.4, tells us that 2x = 27 if and only if x = 7. This means that not only is x = 7 a solution
to 2x = 27 , it is the only solution. Now suppose we change the problem ever so slightly to 2x = 129.
We could use one of the inverse properties of exponentials and logarithms listed in Theorem 6.3 to
write 129 = 2log2 (129) . We’d then have 2x = 2log2 (129) , which means our solution is x = log2 (129).
This makes sense because, after all, the definition of log2 (129) is ‘the exponent we put on 2 to get
129.’ Indeed we could have obtained this solution directly by rewriting the equation 2x = 129 in
its logarithmic form log2 (129) = x. Either way, in order to get a reasonable decimal approximation
to this number, we’d use the change of base formula, Theorem 6.7, to give us something more
calculator friendly,1 say log2 (129) = ln(129)
ln(2) . Another way to arrive at this answer is as follows
2x = 129
ln (2x ) = ln(129)
x ln(2) = ln(129)
ln(129)
x =
ln(2)
Take the natural log of both sides.
Power Rule
‘Taking the natural log’ of both sides is akin to squaring both sides: since f (x) = ln(x) is a function,
as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as
any other non-zero real number and divide it through3 to isolate the variable x. We summarize
below the two common ways to solve exponential equations, motivated by our examples.
Steps for Solving an Equation involving Exponential Functions
1. Isolate the exponential function.
2. (a) If convenient, express both sides with a common base and equate the exponents.
(b) Otherwise, take the natural log of both sides of the equation and use the Power Rule.
Example 6.3.1. Solve the following equations. Check your answer graphically using a calculator.
1. 23x = 161−x
4. 75 =
100
1+3e−2t
2. 2000 = 1000 · 3−0.1t
3. 9 · 3x = 72x
5. 25x = 5x + 6
6.
ex −e−x
2
=5
Solution.
1
You can use natural logs or common logs. We choose natural logs. (In Calculus, you’ll learn these are the most
‘mathy’ of the logarithms.)
2
This is also the ‘if’ part of the statement logb (u) = logb (w) if and only if u = w in Theorem 6.4.
3
Please resist the temptation to divide both
of ln(2). Just like it wouldn’t make sense to
√ sides by ‘ln’ instead
√
divide both sides by the square root symbol ‘ ’ when solving x 2 = 5, it makes no sense to divide by ‘ln’.
6.3 Exponential Equations and Inequalities
449
1−x
1. Since 16 is a power of 2, we can rewrite 23x = 161−x as 23x = 24
. Using properties of
exponents, we get 23x = 24(1−x) . Using the one-to-one property of exponential functions, we
get 3x = 4(1−x) which gives x = 74 . To check graphically, we set f (x) = 23x and g(x) = 161−x
and see that they intersect at x = 47 ≈ 0.5714.
2. We begin solving 2000 = 1000 · 3−0.1t by dividing both sides by 1000 to isolate the exponential
which yields 3−0.1t = 2. Since it is inconvenient to write 2 as a power of 3, we use the natural
log to get ln 3−0.1t = ln(2). Using the Power Rule, we get −0.1t ln(3) = ln(2), so we
10 ln(2)
divide both sides by −0.1 ln(3) to get t = − 0.1ln(2)
ln(3) = − ln(3) . On the calculator, we graph
ln(2)
≈ −6.3093.
f (x) = 2000 and g(x) = 1000 · 3−0.1x and find that they intersect at x = − 10ln(3)
y = f (x) = 23x and
y = g(x) = 161−x
y = f (x) = 2000 and
y = g(x) = 1000 · 3−0.1x
3. We first note that we can rewrite the equation 9·3x = 72x as 32 ·3x = 72x to obtain 3x+2 = 72x .
Since it is not convenientto express both sides as a power of 3 (or 7 for that matter) we use
the natural log: ln 3x+2 = ln 72x . The power rule gives (x + 2) ln(3) = 2x ln(7). Even
though this equation appears very complicated, keep in mind that ln(3) and ln(7) are just
constants. The equation (x + 2) ln(3) = 2x ln(7) is actually a linear equation and as such we
gather all of the terms with x on one side, and the constants on the other. We then divide
both sides by the coefficient of x, which we obtain by factoring.
(x + 2) ln(3)
x ln(3) + 2 ln(3)
2 ln(3)
2 ln(3)
x
=
=
=
=
=
2x ln(7)
2x ln(7)
2x ln(7) − x ln(3)
x(2 ln(7) − ln(3)) Factor.
2 ln(3)
2 ln(7)−ln(3)
Graphing f (x) = 9·3x and g(x) = 72x on the calculator, we see that these two graphs intersect
2 ln(3)
at x = 2 ln(7)−ln(3)
≈ 0.7866.
100
4. Our objective in solving 75 = 1+3e
−2t is to first isolate the exponential. To that end, we
clear denominators and get 75 1 + 3e−2t = 100. From this we get 75 + 225e−2t = 100,
which leads to 225e−2t = 25, and finally, e−2t = 19 . Taking the natural log of both sides
450
Exponential and Logarithmic Functions
gives ln e−2t = ln 19 . Sincenatural log is log base e, ln e−2t = −2t. We can also use
the Power Rule to write ln 19 = − ln(9). Putting these two steps together, we simplify
ln e−2t = ln 91 to −2t = − ln(9). We arrive at our solution, t = ln(9)
which simplifies to
2
t = ln(3). (Can you explain why?) The calculator confirms the graphs of f (x) = 75 and
100
g(x) = 1+3e
−2x intersect at x = ln(3) ≈ 1.099.
y = f (x) = 9 · 3x and
y = g(x) = 72x
y = f (x) = 75 and
100
y = g(x) = 1+3e
−2x
x
5. We start solving 25x = 5x + 6 by rewriting 25 = 52 so that we have 52 = 5x + 6, or
52x = 5x + 6. Even though we have a common base, having two terms on the right hand side
of the equation foils our plan of equating exponents or taking logs. If we stare at this long
enough, we notice that we have three terms with the exponent on one term exactly twice that
of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting u = 5x ,
we have u2 = (5x )2 = 52x so the equation 52x = 5x + 6 becomes u2 = u + 6. Solving this as
u2 − u − 6 = 0 gives u = −2 or u = 3. Since u = 5x , we have 5x = −2 or 5x = 3. Since
5x = −2 has no real solution, (Why not?) we focus on 5x = 3. Since it isn’t convenient to
express 3 as a power of 5, we take natural logs and get ln (5x ) = ln(3) so that x ln(5) = ln(3)
ln(3)
or x = ln(5)
. On the calculator, we see the graphs of f (x) = 25x and g(x) = 5x + 6 intersect
at x =
ln(3)
ln(5)
≈ 0.6826.
x
−x
6. At first, it’s unclear how to proceed with e −e
= 5, besides clearing the denominator to
2
x
−x
−x
obtain e − e = 10. Of course, if we rewrite e = e1x , we see we have another denominator
lurking in the problem: ex − e1x = 10. Clearing this denominator gives us e2x − 1 = 10ex ,
and once again, we have an equation with three terms where the exponent on one term is
exactly twice that of another - a ‘quadratic in disguise.’ If we let u = ex , then u2 = e2x so the
equation e2x − 1 = 10ex can be viewed
as u2 − 1 = 10u. Solving u2 −
√
√10u − 1 = 0, we
√ obtain
x = 5 ± 26. Since 5 − 26 < 0,
by the quadratic formula u = 5 ± √
26. From this, we have
e
√
we get no real
to ex = 5 − 26, but for ex = 5 + 26, we take natural logs to obtain
√ solution
x
−x
x = ln 5 + 26 . If we graph f (x) = e −e
and g(x) = 5, we see that the graphs intersect
2
√ at x = ln 5 + 26 ≈ 2.312
6.3 Exponential Equations and Inequalities
451
x
−x
and
y = f (x) = e −e
2
y = g(x) = 5
y = f (x) = 25x and
y = g(x) = 5x + 6
The authors would be remiss not to mention that Example 6.3.1 still holds great educational
value. Much can be learned about logarithms and exponentials by verifying the solutions obtained
in Example 6.3.1 analytically. For example, to verify our solution to 2000 = 1000 · 3−0.1t , we
ln(2)
substitute t = − 10ln(3)
and obtain
?
2000 = 1000 · 3
10 ln(2)
−0.1 − ln(3)
ln(2)
?
2000 = 1000 · 3 ln(3)
?
Change of Base
?
Inverse Property
2000 = 1000 · 3log3 (2)
2000 = 1000 · 2
X
2000 = 2000
The other solutions can be verified by using a combination of log and inverse properties. Some fall
out quite quickly, while others are more involved. We leave them to the reader.
Since exponential functions are continuous on their domains, the Intermediate Value Theorem 3.1
applies. As with the algebraic functions in Section 5.3, this allows us to solve inequalities using
sign diagrams as demonstrated below.
Example 6.3.2. Solve the following inequalities. Check your answer graphically using a calculator.
1. 2x
2 −3x
− 16 ≥ 0
2.
ex
≤3
ex − 4
3. xe2x < 4x
Solution.
2
1. Since we already have 0 on one side of the inequality, we set r(x) = 2x −3x − 16. The domain
of r is all real numbers, so in order to construct our sign diagram, we seed to find the zeros of
2
2
2
r. Setting r(x) = 0 gives 2x −3x − 16 = 0 or 2x −3x = 16. Since 16 = 24 we have 2x −3x = 24 ,
so by the one-to-one property of exponential functions, x2 − 3x = 4. Solving x2 − 3x − 4 = 0
gives x = 4 and x = −1. From the sign diagram, we see r(x) ≥ 0 on (−∞, −1] ∪ [4, ∞), which
2
corresponds to where the graph of y = r(x) = 2x −3x − 16, is on or above the x-axis.
452
Exponential and Logarithmic Functions
(+) 0 (−) 0 (+)
−1
4
y = r(x) = 2x
2 −3x
− 16
x
2. The first step we need to take to solve exe−4 ≤ 3 is to get 0 on one side of the inequality. To
that end, we subtract 3 from both sides and get a common denominator
ex
ex − 4
≤ 3
ex
−3 ≤ 0
−4
3 (ex − 4)
ex
−
≤ 0 Common denomintors.
ex − 4
ex − 4
12 − 2ex
≤ 0
ex − 4
ex
x
x
We set r(x) = 12−2e
ex −4 and we note that r is undefined when its denominator e − 4 = 0, or
when ex = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4), ∞). To
find the zeros of r, we solve r(x) = 0 and obtain 12 − 2ex = 0. Solving for ex , we find ex = 6,
or x = ln(6). When we build our sign diagram, finding test values may be a little tricky since
we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4
which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the
sign of r (ln(3)) may be very unsettling, remember that eln(3) = 3, so
r (ln(3)) =
12 − 2eln(3)
12 − 2(3)
= −6
=
ln(3)
3−4
e
−4
We determine the signs of r (ln(5)) and r (ln(7)) similarly.5 From the sign diagram, we
find our answer to be (−∞, ln(4)) ∪ [ln(6), ∞). Using the calculator, we see the graph of
x
f (x) = exe−4 is below the graph of g(x) = 3 on (−∞, ln(4)) ∪ (ln(6), ∞), and they intersect
at x = ln(6) ≈ 1.792.
4
This is because the base of ln(x) is e > 1. If the base b were in the interval 0 < b < 1, then logb (x) would
decreasing.
5
We could, of course, use the calculator, but what fun would that be?
6.3 Exponential Equations and Inequalities
453
(−) ‽ (+) 0 (−)
ln(4)
ln(6)
y = f (x) =
ex
ex −4
y = g(x) = 3
3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x − 4x < 0.
We set r(x) = xe2x − 4x and since there are no denominators, even-indexed radicals, or logs,
the domain of r is all real numbers. Setting r(x) = 0 produces xe2x − 4x = 0. We factor
to get x e2x − 4 = 0 which gives x = 0 or e2x − 4 = 0. To solve the latter, we isolate the
exponential and take logs to get 2x = ln(4), or x = ln(4)
2 = ln(2). (Can you explain the last
equality using properties of logs?) As in the previous example,
about
we need to be careful
1
3
6
choosing test values. Since ln(1) = 0, we choose ln 2 , ln 2 and ln(3). Evaluating, we get
r ln
1
2
2 ln( 12 )
1
− 4 ln 12
2 e
2
1
ln 21 eln( 2 ) − 4 ln 12
1
ln 21 eln( 4 ) − 4 ln 21
1
1
1
15
4 ln 2 − 4 ln 2 = − 4
= ln
=
=
=
Power Rule
ln
1
2
Since 21 < 1, ln 12 < 0 and we get r(ln 21 ) is (+), so r(x) < 0 on (0, ln(2)). The calculator
confirms that the graph of f (x) = xe2x is below the graph of g(x) = 4x on these intervals.7
(+) 0 (−) 0 (+)
0
ln(2)
y = f (x) = xe2x and y = g(x) = 4x
6
7
A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method.
Note: ln(2) ≈ 0.693.
454
Exponential and Logarithmic Functions
Example 6.3.3. Recall from Example 6.1.2 that the temperature of coffee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t . When will the coffee be
warmer than 100◦ F?
Solution. We need to find when T (t) > 100, or in other words, we need to solve the inequality
70 + 90e−0.1t > 100. Getting 0 on one side of the inequality, we have 90e−0.1t − 30 > 0, and
we set r(t) = 90e−0.1t − 30. The domain of r is artificially restricted due to the context of the
problem to [0, ∞), so we proceed
to find the zeros of r. Solving 90e−0.1t − 30 = 0 results in
1
1
e−0.1t = 3 so that t = −10 ln 3 which, after a quick application of the Power Rule leaves us with
t = 10 ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1 < 3,
0 = ln(1) < ln(3) so that 10 ln(3) > 0. So we choose t = 0 as a test value in [0, 10 ln(3)). Since
3 < 4, 10 ln(3) < 10 ln(4), so the latter is our choice of a test value for the interval (10 ln(3), ∞).
Our sign diagram is below, and next to it is our graph of t = T (t) from Example 6.1.2 with the
horizontal line y = 100.
y
180
160
140
(+)
0
0 (−)
10 ln(3)
120
y = 100
80
60
H.A. y = 70
40
20
2
4
6
8 10 12 14 16 18 20 t
y = T (t)
In order to interpret what this means in the context of the real world, we need a reasonable
approximation of the number 10 ln(3) ≈ 10.986. This means it takes approximately 11 minutes for
the coffee to cool to 100◦ F. Until then, the coffee is warmer than that.8
We close this section by finding the inverse of a function which is a composition of a rational
function with an exponential function.
5ex
is one-to-one. Find a formula for f −1 (x) and check
ex + 1
your answer graphically using your calculator.
Solution. We start by writing y = f (x), and interchange the roles of x and y. To solve for y, we
first clear denominators and then isolate the exponential function.
Example 6.3.4. The function f (x) =
8
Critics may point out that since we needed to use the calculator to interpret our answer 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
455
Switch x and y
x (ey + 1) = 5ey
xey + x = 5ey
x = 5ey − xey
x = ey (5 − x)
x
ey =
5−x
x
y
ln (e ) = ln
5−x
x
y = ln
5−x
x
We claim f −1 (x) = ln 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
on the same set of axes and observe the symmetry about the line y = x.
and y = g(x) = ln 5−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
456
6.3.1
Exponential and Logarithmic Functions
Exercises
In Exercises 1 - 33, solve the equation analytically.
1. 24x = 8
4. 42x =
2. 3(x−1) = 27
1
2
5. 8x =
7. 37x = 814−2x
1
128
8. 9 · 37x =
10. 5−x = 2
11. 5x = −2
13. (1.005)12x = 3
16. 500 1 − e2x = 250
14. e−5730k =
100ex
= 50
ex + 2
x
22. 25 45 = 10
19.
3. 52x−1 = 125
6. 2(x
1 2x
9
=1
9. 32x = 5
12. 3(x−1) = 29
1
2
15. 2000e0.1t = 4000
17. 70 + 90e−0.1t = 75
20.
3 −x)
5000
= 2500
1 + 2e−3t
18. 30 − 6e−0.1x = 20
21.
150
= 75
1 + 29e−0.8t
24. 7e2x = 28e−6x
23. e2x = 2ex
1 (x+5)
2
25. 3(x−1) = 2x
26. 3(x−1) =
27. 73+7x = 34−2x
28. e2x − 3ex − 10 = 0
29. e2x = ex + 6
30. 4x + 2x = 12
31. ex − 3e−x = 2
32. ex + 15e−x = 8
33. 3x + 25 · 3−x = 10
In Exercises 34 - 39, solve the inequality analytically.
35. 1000 (1.005)12t ≥ 3000
x
37. 25 54 ≥ 10
34. ex > 53
36. 2(x
38.
3 −x)
<1
150
≤ 130
1 + 29e−0.8t
39. 70 + 90e−0.1t ≤ 75
In Exercises 40 - 45, use your calculator to help you solve the equation or inequality.
√
40. 2x = x2
41. ex = ln(x) + 5
42. e
x
43. e−x − xe−x ≥ 0
44. 3(x−1) < 2x
45. ex < x3 − x
=x+1
46. Since f (x) = ln(x) is a strictly increasing function, if 0 < a < b then ln(a) < ln(b). Use this
fact to solve the inequality e(3x−1) > 6 without a sign diagram. Use this technique to solve
the inequalities in Exercises 34 - 39. (NOTE: Isolate the exponential function first!)
47. Compute the inverse of f (x) =
ex − e−x
. State the domain and range of both f and f −1 .
2
6.3 Exponential Equations and Inequalities
5ex
48. In Example 6.3.4, we found that the inverse of f (x) = x
was f −1 (x) = ln
e +1
we left a few loose ends for you to tie up.
457
x
5−x
but
(a) Show that 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 .
(b) Find the range of f by finding the domain of f −1 .
5x
(c) Let g(x) =
and h(x) = ex . Show that f = g ◦ h and that (g ◦ h)−1 = h−1 ◦ g −1 .
x+1
(We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a
specific example of the property.)
49. With the help of your classmates, solve the inequality ex > xn for a variety of natural
numbers n. What might you conjecture about the “speed” at which f (x) = ex grows versus
any polynomial?
458
6.3.2
Exponential and Logarithmic Functions
Answers
1. x =
3
4
4. x = − 14
7. x =
16
15
ln(2)
10. x = − ln(5)
ln(3)
12 ln(1.005)
16. x =
1
2
1
2
22. x =
1
3
6. x = −1, 0, 1
2
8. x = − 11
9. x =
14. k =
ln( 12 )
−5730
= 10 ln
ln( 52 )
ln( 54 )
=
1
5730
ln(2)
17. t =
3
5
21. t =
=
1
4
=
ln(3)+5 ln( 12 )
ln(3)−ln( 12 )
1
4
=
12. x =
ln(29)+ln(3)
ln(3)
15. t =
ln(2)
0.1
= 10 ln(18)
1
ln( 29
)
−0.8
=
5
4
ln(29)
ln(2)
25. x =
ln(3)
ln(3)−ln(2)
ln(3)−5 ln(2)
ln(3)+ln(2)
27. x =
4 ln(3)−3 ln(7)
7 ln(7)+2 ln(3)
28. x = ln(5)
29. x = ln(3)
31. x = ln(3)
32. x = ln(3), ln(5)
34. (ln(53), ∞)
36. (−∞, −1) ∪ (0, 1)
2
ln( 377
)
38. −∞, −0.8 = −∞, 45 ln
= 10 ln(2)
23. x = ln(2)
ln(2)−ln(5)
ln(4)−ln(5)
1
ln( 18
)
−0.1
ln(5)
2 ln(3)
19. x = ln(2)
ln(2)
24. x = − 18 ln
26. x =
5. x = − 73
= − 21 ln(2)
5
3
18. x = −10 ln
20. t =
3. x = 2
11. No solution.
13. x =
ln
2. x = 4
377
2
30. x =
ln(3)
ln(2)
ln(5)
33. x = ln(3)
h
ln(3)
35. 12 ln(1.005)
,∞
i
ln( 52 )
37. −∞, ln 4 = −∞, ln(2)−ln(5)
ln(4)−ln(5)
(5)
1
ln( 18 )
39. −0.1
, ∞ = [10 ln(18), ∞)
40. x ≈ −0.76666, x = 2, x = 4
41. x ≈ 0.01866, x ≈ 1.7115
42. x = 0
43. (−∞, 1]
44. ≈ (−∞, 2.7095)
45. ≈ (2.3217, 4.3717)
46. x > 13 (ln(6) + 1)
√
47. f −1 = ln x + x2 + 1 . Both f and f −1 have domain (−∞, ∞) and range (−∞, ∞).
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