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6.2 Properties of Logarithms 437
6.2 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
eliminates the pathologies discussed in Section 5.3 when, for example, we simplified
x^2 /^3
and obtained |x| instead of what we had expected from the arithmetic in the exponents, x^1 = 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^ = bubw
- Quotient Rule: f (u − w) =
f (u) f (w)
. In other words, bu−w^ =
bu bw
- 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
438 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)
- Quotient Rule: g
( (^) u w
= g(u) − g(w). In other words, logb
( (^) u w
= logb(u) − logb(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 = bcbd^ = 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.
- log 2
x
- log 0. 1
10 x^2
- ln
ex
- log 3
100 x^2 yz^5
- log 117
x^2 − 4
Solution.
- To expand log 2
x
, we use the Quotient Rule identifying u = 8 and w = x and simplify. (^1) 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.
440 Exponential and Logarithmic Functions
Rule^2 , and we keep in mind that the common log is log base 10.
log 3
100 x^2 yz^5 = log
100 x^2 yz^5
= 13 log
100 x^2 yz^5
Power Rule
= (^13)
[
log
100 x^2
− log
yz^5
)]
Quotient Rule = 13 log
100 x^2
− 13 log
yz^5
[
log(100) + log
x^2
)]
[
log(y) + log
z^5
)]
Product Rule = 13 log(100) + 13 log
x^2
− 13 log(y) − 13 log
z^5
= 13 log(100) + 23 log(x) − 13 log(y) − 53 log(z) Power Rule = 23 + 23 log(x) − 13 log(y) − 53 log(z) Since 10^2 = 100 = 23 log(x) − 13 log(y) − 53 log(z) + (^23)
- At first it seems as if we have no means of simplifying log 117
x^2 − 4
, since none of the properties of logs addresses the issue of expanding a difference inside the logarithm. However, we may factor x^2 − 4 = (x + 2)(x − 2) thereby introducing a product which gives us license to use the Product Rule.
log 117
x^2 − 4
= log 117 [(x + 2)(x − 2)] Factor = log 117 (x + 2) + log 117 (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 log 0. 1
10 x^2
, 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) = log 117
x^2 − 4
and g(x) = log 117 (x + 2) + log 117 (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 like log 117
x^2 − 4
= log 117
x^2
− log 117 (4), which simply isn’t true, in general. The unwritten^3 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.
- log 3 (x − 1) − log 3 (x + 1) 2. log(x) + 2 log(y) − log(z)
- 4 log 2 (x) + 3 4. − ln(x) − (^12)
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.
- The difference of logarithms requires the Quotient Rule: log 3 (x−1)−log 3 (x+1) = log 3
x− 1 x+
- 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) Power Rule
= log
xy^2
− log(z) Product Rule
= log
xy^2 z
Quotient Rule
- We can certainly get started rewriting 4 log 2 (x) + 3 by applying the Power Rule to 4 log 2 (x) to obtain log 2
x^4
, 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 = log 2
, so we get
4 log 2 (x) + 3 = log 2
x^4
x^4
Since 3 = log 2
= log 2
x^4
8 x^4
Product Rule
(^3) The authors relish the irony involved in writing what follows.
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.
- 3^2 to base 10 2. 2x^ to base e
- log 4 (5) to base e 4. ln(x) to base 10
Solution.
- We apply the Change of Base formula with a = 3 and b = 10 to obtain 3^2 = 102 log(3). Typing the latter in the calculator produces an answer of 9 as required.
- 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.
444 Exponential and Logarithmic Functions
- Applying the change of base with a = 4 and b = e leads us to write log 4 (5) = ln(5)ln(4). Evaluating this in the calculator gives ln(5)ln(4) ≈ 1 .16. How do we check this really is the value of log 4 (5)? By definition, log 4 (5) is the exponent we put on 4 to get 5. The calculator confirms this.^5
- We write ln(x) = loge(x) = log(log(xe)). We graph both f (x) = ln(x) and g(x) = log(log(xe)) and find both graphs appear to be identical.
y = f (x) = ln(x) and y = g(x) = log(log(xe))
(^5) Which means if it is lying to us about the first answer it gave us, at least it is being consistent.
446 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.
- 7x−^1 to base e 31. log 3 (x + 2) to base 10
)x to base e 33. log(x^2 + 1) to base e
In Exercises 34 - 39, use the appropriate change of base formula to approximate the logarithm.
- log 3 (12) 35. log 5 (80) 36. log 6 (72)
- log 4
- log 3 5 (1000) 39. log 2 3
- Compare and contrast the graphs of y = ln(x^2 ) and y = 2 ln(x).
- Prove the Quotient Rule and Power Rule for Logarithms.
- 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)
(c) logb
x y
logb(x) logb(y)
- 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−]
[HA]
= [H+]
[A−]
[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 pH = pKa + log
[A−]
[HA]
- Research the history of logarithms including the origin of the word ‘logarithm’ itself. Why is the abbreviation of natural log ‘ln’ and not ‘nl’?
- 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 447
6.2.2 Answers
- 3 ln(x) + 2 ln(y) 2. 7 − log 2 (x^2 + 4)
- 3 log 5 (z) − 6 4. log(1.23) + 37
- 12 ln(z) − ln(x) − ln(y) 6. log 5 (x − 5) + log 5 (x + 5)
- 3 log√ 2 (x) + 4 8. −2 + log 13 (x) + log 13 (y − 2) + log 13 (y^2 + 2y + 4)
- 3 + 3 log(x) + 5 log(y) 10. 2 log 3 (x) − 4 − 4 log 3 (y)
- 14 ln(x) + 14 ln(y) − 14 − 14 ln(z) 12. 12 − 12 log 6 (x) − 4 log 6 (y)
- 53 + log(x) + 12 log(y) 14. −2 + 23 log 1 2 (x) − log 1 2 (y) − 12 log 1 2 (z)
- 13 ln(x) − ln(10) − 12 ln(y) − 12 ln(z) 16. ln(x^4 y^2 )
- log 2
( (^) xy z
- log 3
x y^2
- log 3
x y^2 z
- ln
x^2 y^3 z^4
- log
( (^) x√y √ (^3) z
- ln
3
z xy
- log 5
( (^) x 125
- log
x
- log 7
x(x−3) 49
- ln (x
e) 27. log 2
x^3 /^2
- log 2
x
x − 1
- log 2
x x− 1
- 7x−^1 = e(x−1) ln(7)^ 31. log 3 (x + 2) = log(log(3)x+2)
3
)x = ex^ ln(^
(^23) )
- log(x^2 + 1) = ln(x
(^2) +1) ln(10)
- log 3 (12) ≈ 2. 26186 35. log 5 (80) ≈ 2. 72271
- log 6 (72) ≈ 2. 38685 37. log 4
10
- log 3 5 (1000) ≈ − 13. 52273 39. log 2 3
6.3 Exponential Equations and Inequalities 449
- Since 16 is a power of 2, we can rewrite 2^3 x^ = 16^1 −x^ as 2^3 x^ =
) 1 −x
. Using properties of exponents, we get 2^3 x^ = 24(1−x). Using the one-to-one property of exponential functions, we get 3x = 4(1−x) which gives x = 47. To check graphically, we set f (x) = 2^3 x^ and g(x) = 16^1 −x and see that they intersect at x = 47 ≈ 0 .5714.
- We begin solving 2000 = 1000 · 3 −^0.^1 t^ by dividing both sides by 1000 to isolate the exponential which yields 3−^0.^1 t^ = 2. Since it is inconvenient to write 2 as a power of 3, we use the natural log to get ln
3 −^0.^1 t
= ln(2). Using the Power Rule, we get − 0. 1 t ln(3) = ln(2), so we divide both sides by − 0 .1 ln(3) to get t = − (^0) .ln(2)1 ln(3) = − 10 ln(2)ln(3). On the calculator, we graph f (x) = 2000 and g(x) = 1000 · 3 −^0.^1 x^ and find that they intersect at x = − 10 ln(2)ln(3) ≈ − 6 .3093.
y = f (x) = 2^3 x^ and y = f (x) = 2000 and y = g(x) = 16^1 −x^ y = g(x) = 1000 · 3 −^0.^1 x
- We first note that we can rewrite the equation 9· 3 x^ = 7^2 x^ as 3^2 · 3 x^ = 7^2 x^ to obtain 3x+2^ = 7^2 x. Since it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use the natural log: ln
3 x+
= ln
72 x
. 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) = 2 x ln(7) x ln(3) + 2 ln(3) = 2 x ln(7) 2 ln(3) = 2 x ln(7) − x ln(3) 2 ln(3) = x(2 ln(7) − ln(3)) Factor. x = (^) 2 ln(7)2 ln(3)−ln(3)
Graphing f (x) = 9· 3 x^ and g(x) = 7^2 x^ on the calculator, we see that these two graphs intersect at x = (^) 2 ln(7)2 ln(3)−ln(3) ≈ 0 .7866.
- Our objective in solving 75 = (^) 1+3^100 e− 2 t is to first isolate the exponential. To that end, we clear denominators and get 75
1 + 3e−^2 t
= 100. From this we get 75 + 225e−^2 t^ = 100, which leads to 225e−^2 t^ = 25, and finally, e−^2 t^ = 19. Taking the natural log of both sides
450 Exponential and Logarithmic Functions
gives ln
e−^2 t
= ln
9
. Since natural log is log base e, ln
e−^2 t
= − 2 t. We can also use the Power Rule to write ln
9
= − ln(9). Putting these two steps together, we simplify ln
e−^2 t
= ln
9
to − 2 t = − ln(9). We arrive at our solution, t = ln(9) 2 which simplifies to t = ln(3). (Can you explain why?) The calculator confirms the graphs of f (x) = 75 and g(x) = (^) 1+3^100 e− 2 x intersect at x = ln(3) ≈ 1 .099.
y = f (x) = 9 · 3 x^ and y = f (x) = 75 and y = g(x) = 7^2 x^ y = g(x) = (^) 1+3^100 e− 2 x
- We start solving 25x^ = 5x^ + 6 by rewriting 25 = 5^2 so that we have
)x = 5x^ + 6, or 52 x^ = 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 u^2 = (5x)^2 = 5^2 x^ so the equation 5^2 x^ = 5x^ + 6 becomes u^2 = u + 6. Solving this as u^2 − u − 6 = 0 gives u = −2 or u = 3. Since u = 5x, we have 5x^ = −2 or 5x^ = 3. Since 5 x^ = −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) or x = ln(3)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.
- At first, it’s unclear how to proceed with e x−e−x 2 = 5, besides clearing the denominator to obtain ex^ − e−x^ = 10. Of course, if we rewrite e−x^ = (^) e^1 x , we see we have another denominator lurking in the problem: ex^ − (^) e^1 x = 10. Clearing this denominator gives us e^2 x^ − 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 u^2 = e^2 x^ so the equation e^2 x^ − 1 = 10ex^ can be viewed as u^2 − 1 = 10u. Solving u^2 − 10 u − 1 = 0, we obtain by the quadratic formula u = 5 ±
- From this, we have ex^ = 5 ±
- Since 5 −
we get no real solution to ex^ = 5 −
26, but for ex^ = 5 +
26, we take natural logs to obtain x = ln
. If we graph f (x) = e x−e−x 2 and^ g(x) = 5, we see that the graphs intersect at x = ln
452 Exponential and Logarithmic Functions
y = r(x) = 2x (^2) − 3 x − 16
- The first step we need to take to solve e x ex− 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
ex ex^ − 4
ex ex^ − 4
3 (ex^ − 4) ex^ − 4
≤ 0 Common denomintors.
12 − 2 ex ex^ − 4
We set r(x) = 12 −^2 e x ex− 4 and we note that^ r^ is undefined when its denominator^ e
x (^) − 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 − 2 ex^ = 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 increasing^4 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 − 2 eln(3) eln(3)^ − 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 f (x) = e x ex− 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
ln(4)
ln(6)
y = f (x) = e x ex− 4 y = g(x) = 3
- As before, we start solving xe^2 x^ < 4 x by getting 0 on one side of the inequality, xe^2 x^ − 4 x < 0. We set r(x) = xe^2 x^ − 4 x and since there are no denominators, even-indexed radicals, or logs, the domain of r is all real numbers. Setting r(x) = 0 produces xe^2 x^ − 4 x = 0. We factor to get x
e^2 x^ − 4
= 0 which gives x = 0 or e^2 x^ − 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, we need to be careful about choosing test values. Since ln(1) = 0, we choose ln
2
, ln
2
and ln(3). Evaluating,^6 we get
r
ln
2
= ln
2
e2 ln(^
− 4 ln
2
= ln
2
eln(^
1 2 )
2 − 4 ln
2
Power Rule = ln
2
eln(^
− 4 ln
2
= 14 ln
2
− 4 ln
2
= − 154 ln
2
Since 12 < 1, ln
2
< 0 and we get r(ln
2
) is (+), so r(x) < 0 on (0, ln(2)). The calculator confirms that the graph of f (x) = xe^2 x^ is below the graph of g(x) = 4x on these intervals.^7
ln(2)
y = f (x) = xe^2 x^ and y = g(x) = 4x
(^6) A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. (^7) Note: ln(2) ≈ 0 .693.
6.3 Exponential Equations and Inequalities 455
y = 5 ex ex^ + 1
x = 5 ey ey^ + 1
Switch x and y
x (ey^ + 1) = 5 ey xey^ + x = 5 ey x = 5 ey^ − xey x = ey(5 − x) ey^ = x 5 − x
ln (ey) = ln
x 5 − x
y = ln
x 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) = 5 e x ex+ 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.
y = f (x) = 5 e x ex+1 and^ y^ =^ g(x) = ln
x 5 −x
456 Exponential and Logarithmic Functions
6.3.1 Exercises
In Exercises 1 - 33, solve the equation analytically.
- 2^4 x^ = 8 2. 3(x−1)^ = 27 3. 5^2 x−^1 = 125
- 4^2 x^ = 12 5. 8x^ = 1281 6. 2(x (^3) −x) = 1
- 3^7 x^ = 81^4 −^2 x^ 8. 9 · 37 x^ =
9
) 2 x
- 3^2 x^ = 5
- 5−x^ = 2 11. 5x^ = − 2 12. 3(x−1)^ = 29
- (1.005)^12 x^ = 3 14. e−^5730 k^ = 12 15. 2000e^0.^1 t^ = 4000
- 500
1 − e^2 x
= 250 17. 70 + 90e−^0.^1 t^ = 75 18. 30 − 6 e−^0.^1 x^ = 20
100 ex ex^ + 2
= 50 20.^
1 + 2e−^3 t^
1 + 29e−^0.^8 t^
5
)x = 10 23. e^2 x^ = 2ex^ 24. 7e^2 x^ = 28e−^6 x
- 3(x−1)^ = 2x^ 26. 3(x−1)^ =
2
)(x+5)
- 73+7x^ = 3^4 −^2 x
- e^2 x^ − 3 ex^ − 10 = 0 29. e^2 x^ = ex^ + 6 30. 4x^ + 2x^ = 12
- ex^ − 3 e−x^ = 2 32. ex^ + 15e−x^ = 8 33. 3x^ + 25 · 3 −x^ = 10
In Exercises 34 - 39, solve the inequality analytically.
- ex^ > 53 35. 1000 (1.005)^12 t^ ≥ 3000
- 2(x (^3) −x) < 1 37. 25
5
)x ≥ 10
1 + 29e−^0.^8 t^
≤ 130 39. 70 + 90e−^0.^1 t^ ≤ 75
In Exercises 40 - 45, use your calculator to help you solve the equation or inequality.
- 2x^ = x^2 41. ex^ = ln(x) + 5 42. e
√x = x + 1
- e−x^ − xe−x^ ≥ 0 44. 3(x−1)^ < 2 x^ 45. ex^ < x^3 − x
- 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!)
- Compute the inverse of f (x) =
ex^ − e−x 2
. State the domain and range of both f and f −^1.
