163

Section 4.3 - Logarithmic Functions Objective 1: Definition of a Logarithmic Function Let’s work with solving some exponential equations. Recall exponential functions are one-to-one with a domain of (– ∞, ∞) and a range of (0, ∞). We want to use these facts to solve the following. Solve the following: Ex. 1a 4x = 16 Ex. 1b (1/3)x = 243 Ex. 1c 5x = 625 Ex. 1d 2x =

€

18

Ex. 1e 6x = 0 Ex. 1f 5x = – 25 Solution: a) Since 16 = 42, then x = 2. b) Since (1/3)x = 3 – x = 243 and 243 = 35, then 3 – x = 35. Thus, – x = 5 which means x = – 5. c) Since 625 = 54, then x = 4. d) Since

€

18

=

€

123 = 2 – 3, then x = – 3.

e) Recall that the range of the exponential function is (0, ∞). Thus, 6x = 0 has no solution. f) Recall that the range of the exponential function is (0, ∞). Thus, 5x = – 25 has no solution. Now, suppose we want to solve 4x = 13. The problem with 13 is that it is not a perfect power of four. We can make a conjecture that x should be between 1 and 2 since 41 = 4 and 42 = 16 and 13 is between 4 and 16. We could try some values for x Try x = 1.5, we get 41.5 = 8 (too low) Try x = 1.75, we get 41.75 ≈ 11.3137... (too low) Try x = 1.875, we get 41.875 ≈ 13.4543... (too high) Try x = 1.8125, we get 41.8125 ≈ 12.3376... (too low) Etc. It will take quite a while to find x = 1.850219859... is the answer. As we can see, this process is far too time consuming. We need to develop a better method for solving this equation. Recall that if a function is one-to-one, then it has an inverse function. Since an exponential function is one-to-one, then it has an inverse function. We call this inverse function a logarithmic function.

164

Definition of a Logarithmic Function For x > 0 and a > 0 and a ≠ 1, then y = f(x) = loga (x) if and only if x = ay. This is called the logarithmic function with base a and x is called the argument. Note, the domain of loga (x) is (0, ∞) and the range is (– ∞, ∞).

Think of the logarithmic functions as asking the question: "By what power f(x) do we have to raise the base a to get x?"

Argument = BasePower Power = logBase(Argument)

Rewrite each exponential function as an logarithmic function: Ex. 2a 3.53 = r Ex. 2b ew = 8 Ex. 2c b6 = 56 Solution: a) Think "what power 3 do we have to raise 3.5 to get r?" 3 = log3.5(r) b) Think "what power w do we have to raise e to get 8?" w = loge(8) c) Think "what power 6 do we have to raise b to get 56?" 6 = logb(56) Rewrite each logarithmic function as an exponential function:

Ex. 3a log12(144) = p Ex. 3b logc(1000) = 3 Ex. 3c log4(x) = – 3 Solution: a) With log12(144) = p, 12 is the base, p is the power and 144 is the “answer”. Thus, 12p = 144.

b) With logc(1000) = 3, c is the base, 3 is the power and 1000 is the “answer”. Thus, c3 = 1000.

c) With log4(x) = – 3, 4 is the base, – 3 is the power and x is the “answer”. Thus, 4 – 3 = x. Objective 2: Evaluating Logarithmic Functions

Find the following: Ex. 4a log10(100) Ex. 4b log2(

€

25 ) Ex. 4c log3(

€

181

) Ex. 4d log5 (5) Ex. 4e log7 (1) Ex. 4f log2 (0)

165

Solution: a) For log10(100), ask the question: "By what power f(x) do we have to raise the base 10 to get 100?" The answer is 2. Thus, since 102 = 100, then log10(100) = 2.

b) For log2(

€

25 ), ask the question: "By what power f(x) do we have to raise the base 2 to get

€

25 = 21/5?" The answer is 1/5. Thus, since 21/5 =

€

25 , then log2(

€

25 ) = 1/5.

c) For log3(

€

181

), ask the question: "By what power f(x) do we have

to raise the base 3 to get

€

181

?" Since

€

181

= 81 – 1 = (34) – 1 = 3 – 4,

the answer is – 4. Thus, since 3 – 4 =

€

181

, then log3(

€

181

) = – 4.

d) For log5(5), ask the question: "By what power f(x) do we have to raise the base 5 to get 5?" The answer is 1. Thus, since 51 = 5, then log5(5) = 1.

e) For log7(1), ask the question: "By what power f(x) do we have to raise the base 7 to get 1?" The answer is 0. Thus, since 70 = 1, then log7(1) = 0.

f) The domain of the logarithmic function is (0, ∞). Thus, log2 (0) is undefined. Solve the following: Ex. 5a logx(

€

1681 ) = 4 Ex. 5b log8 x =

€

43

Ex. 5c log36

€

65 = x Solution: a) logx(

€

1681 ) = 4 (write in exponential form)

x4 =

€

1681

(take the fourth root of both sides)

x = ±

€

1681

4 = ±

€

23

, but the base cannot be negative, so

x =

€

23

166

b) log8 x =

€

43

(write in exponential form) 84/3 = x (write as a radical) (

€

83 )4 = x (simplify) (2)4 = x x = 16 c) log36

€

65 = x (write in exponential form) 36x =

€

65 (write each side with a base of 6) (62)x = 61/5 (simplify) 62x = 61/5 (exponents are equal since the function is 1-1) 2x =

€

15

(solve)

x =

€

110

Objective 3: Graphing Logarithmic Functions. Since logarithmic functions are inverse functions of exponential function, then the graph of the logarithmic function and the graph of the exponential function should be symmetrical to the line y = x. a > 1 0 < a < 1

Let’s verify these results by making a table of values and plotting some points.

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

ax

loga(x)

ax

loga(x)

167

Make a table of values and graph the following: Ex. 6a f(x) = log2(x) Ex. 6b g(x) = log3(x) Ex. 6c h(x) = log4(x) Solution:

a) x log2(x) b) x log3(x) c) x log4(x)

€

18 – 3

€

127 – 3

€

164 – 3

€

14 – 2

€

19 – 2

€

116 – 2

€

12 – 1

€

13 – 1

€

14 – 1

1 0 1 0 1 0 2 1 3 1 4 1 4 2 9 2 16 2 8 3 27 3 64 3

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

Notice that these functions have the same basic shape though as the base gets bigger, the graph gets "steeper." In general, if a > 1, loga(x) will have this shape.

a) f(x) = log2(x) b) g(x) = log3(x)

c) h(x) = log4(x)

168

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

Make a table of values and graph the following: Ex. 7a f(x) = log1/2(x) Ex. 7b g(x) = log1/3(x) Ex. 7c h(x) = log1/4(x) Solution:

a) x log1/2(x) b) x log1/3(x) c) x log1/4(x) 8 – 3 27 – 3 27 – 3 4 – 2 9 – 2 9 – 2 2 – 1 3 – 1 3 – 1 1 0 1 0 1 0

€

12 1

€

13 1

€

13 1

€

14 2

€

19 2

€

19 2

€

18 3

€

127 3

€

127 3

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

Notice that these functions have the same basic shape though as the base gets bigger, the graph gets "steeper." In general, if 0 < a < 1, loga(x) will have this shape.

a) f(x) = log1/2(x) b) g(x) = log1/3(x)

c) h(x) = log1/4(x)

169

Now, let's summarize what we know about logarithmic functions: The graph of loga(x), a > 1 The graph of loga(x), 0 < a < 1

Domain: (0, ∞) Domain: (0, ∞)

Range: (– ∞, ∞) Range: (– ∞, ∞)

X-Intercept: (1, 0) X-intercept: (1, 0)

No y-intercept No y-intercept

Increasing Decreasing

x = 0 is a Vertical Asymptote x = 0 is a Vertical Asymptote

One-to-One One-to-One

The Common Logarithmic Function log10 (x) is called the common logarithmic function. It is usually denoted as log (x). Thus, if there is no base written, we will assume that it is the common log. The log key on a scientific calculator is the common logarithmic function. Thus, y = log(x) if and only if 10y = x. Evaluate the following using a calculator: Ex. 8a log(376) Ex. 8b log(1.6 × 107) Ex. 8c log(0.0075) Solution: a) log(376) = 2.5751878... ≈ 2.5752 b) log(1.6 × 107) = log(16,000,000) = 7.20411998... ≈ 7.2041 c) log(0.0075) = – 2.12493873... ≈ – 2.1249

(1, 0)

(1, 0)

170

The Natural Logarithmic Function loge (x) is called the natural logarithmic function. It is usually denoted as ln(x). Thus, if there it is written ln, we will assume that it is the natural log. The ln key on a scientific calculator is the natural logarithmic function. Thus, y = ln(x) if and only if ey = x. Evaluate the following using a calculator: Ex. 9a ln(95) Ex. 9b ln(0.08) Ex. 9c ln(4.5) Solution: a) ln(95) = 4.55387689... ≈ 4.5539 b) ln(0.08) = – 2.525728644... ≈ – 2.5257 c) ln(4.5) = 1.504077396... ≈ 1.5041 Since both e and 10 are greater than 1, their graphs will like the first type of log graphs we examined: f(x) = log(x) f(x) = ln(x)

Graph the following Logarithmic Functions: Ex. 10a g(x) = ln(x – 2) + 3 Ex. 10b h(x) = – 2•log(x + 3) Solution: Solution: a) This is the graph of b) This is the graph of the common the natural log function log function stretched by 2, reflected shifted up 3 and right 2. across the x-axis and shifted 3 left.

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

0

1

2

3

4

-1 0 1 2 3 4 5 6 7

171

Objective 4: Work with the Properties of Logarithms

Establish the following properties: Ex. 11a loga(1) = 0 Ex. 11b loga(a) = 1 Solution: a) Let y = loga(1) (rewrite as an exponential equation) ay = 1 (replace 1 by a0 since a > 0 and a ≠ 1) ay = a0 (exponential functions are one-to-one) y = 0 (replace y with loga(1)) loga(1) = 0 b) Let z = loga(a) (rewrite as an exponential equation) az = a (replace a by a1 since a > 0 and a ≠ 1) az = a1 (exponential functions are one-to-one) z = 1 (replace y with loga(a)) loga(a) = 1 Thus, loga(1) = 0 and loga(a) = 1.

Theorem Let M > 0, a > 0, a ≠ 1, and r be any real number. Then 1) aloga(M) = M and 2) loga(ar) = r Proof: 1) Let y = loga(M) (rewrite as an exponential equation) ay = M (replace y by loga(M)) aloga(M) = M

-1

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 -4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

172

2) Let z = ar (rewrite as a logarithmic equation) loga(z) = r (replace z by ar) loga(ar) = r Notice that these two properties reinforce the notion that f(x) = ax and f – 1(x) = loga(x) are inverse functions. Evaluate the following: Ex. 12a log3(3π) Ex. 12b 5log5(e) Ex. 12c ln(ert) Solution: a) log3(3π) = π by property #2 b) 5log5(e) = e by property #1 c) ln(ert) = rt by property #2 Theorem Let M > 0, N > 0, a > 0, a ≠ 1, and r be any real number. Then 3) loga(M•N) = loga(M) + loga(N) The log of the product is the sum of the logs of the factors. 4) loga(

€

MN ) = loga(M) – loga(N)

The log of the quotient is the log of the numerator minus the log of the denominator. 5) loga(Mr) = r loga(M) The log of a power is the exponent times the log of the base. 6) ax = ex ln(a) Proof: Since exponential functions are one-to-one and M & N > 0, there are unique real numbers A and B such that M = aA and N = aB. This implies that A = loga(M) and B = loga(N) Thus, 3) loga(M•N) (replace M by aA and N by aB) = loga(aA•aB) (add the exponents, keep the same base) = loga(aA + B) (use property #2) = A + B (replace A by loga(M) and B by loga(N)) = loga(M) + loga(N)

173

4) loga(

€

MN ) (replace M by aA and N by aB)

= loga(

€

aA

aB ) (subtract the exponents, keep the same base)

= loga(aA – B) (use property #2) = A – B (replace A by loga(M) and B by loga(N)) = loga(M) – loga(N) 5) loga(Mr) (replace M by aA) = loga([aA]r) (multiply the exponents, keep the same base) = loga(aAr) (use property #2) = Ar = rA (replace A by loga(M)) = r loga(M) 6) Consider ln(ax), by property #5, ln(ax) = x ln(a) (exponential functions are one-to-one) eln(ax) = ex ln(a) (eln(ax) = ax by property #1) ax = ex ln(a) Evaluate the following using the properties of logarithms: Ex. 13 Given logb(6) ≈ 1.2 and logb(11) ≈ 1.6, find:

a) logb(66) b) logb(

€

611

) c) logb(6b2) Solution: We need to our properties of logs to rewrite each problem in terms of logb(6) and logb(11). a) logb(66) = logb(6•11) (apply property #3) = logb(6) + logb(11) (replace logb(6) by 1.2 & logb(11) by 1.6) ≈ 1.2 + 1.6 = 2.8

b) logb(

€

611

) = logb(

€

611⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1/ 2

) (apply property #5)

=

€

12

• logb(

€

611

) (apply property #5)

= 12 [ logb(6) – logb(11)] (replace logb(6) by 1.2 & logb(11) by 1.6)

≈ 12 [1.2 – 1.6] = – 0.2

174

c) logb(6b2) (apply property #3) = logb(6) + logb(b2) (apply property #5) = logb(6) + 2•logb(b) (logb(b) = 1) = logb(6) + 2•1 (replace logb(6) by 1.2) = 1.2 + 2•1 = 3.2 We can use the properties of logarithms to rewrite the logarithm of a product and/or quotient as the sum and difference of logarithms. This is an extremely important tool for Calculus. Write as the sum and/or difference of logarithms. Express all powers as factors. Assume the variables are positive:

Ex. 14a ln

€

x2y3

z⎛

⎝ ⎜

⎞

⎠ ⎟ Ex. 14b ln(

€

5xx2 +1

)

Ex. 14c loga [(

€

5x2 x2 +164

4(x−5) )2] , x > 5

Solution: Again, we need to use our properties of logs:

a) ln

€

x2y3

z⎛

⎝ ⎜

⎞

⎠ ⎟ (apply property #4)

= ln(x2y3) – ln(z) (apply property #3) = ln(x2) + ln(y3) – ln(z) (apply property #5) = 2ln(x) + 3ln(y) – ln(z) b) ln(

€

5xx2 +1

) (apply property #4)

= ln(5x) – ln[(x2 + 1)1/2

] (apply property #5) = ln(5x) –

12 •ln(x2 + 1) (apply property #3)

= ln(5) + ln(x) – 12 ln(x2 + 1)

c) loga [(

€

5x2 x2 +164

4(x−5) )2] (apply property #5)

= 2loga (

€

5x2 x2 +164

4(x−5) ) (apply property #4)

175

= 2{loga(5x2

€

x2 +164 ) – loga(4(x – 5))} (apply property #3) = 2{loga(5) + loga(x2) + loga[(x2 + 16)1/4] – {loga(4) + loga(x – 5)}} (distribute the negative sign) = 2{loga(5) + loga(x2) + loga[(x2 + 16)1/4] – loga(4) – loga(x – 5)} (distribute the 2) = 2loga(5) + 2loga(x2) + 2loga[(x2 + 16)1/4] – 2loga(4) – 2loga(x – 5) (apply property #5 and simplify) = 2loga(5) + 2•2loga(x) + 2•

€

14

loga(x2 + 16) – 2loga(4) – 2loga(x – 5)

= 2loga(5) + 4loga(x) +

€

12

loga(x2 + 16) – 2loga(4) – 2loga(x – 5) Another tool we need to solve equations involving logarithmic expressions is to take the sum and/or difference of logarithms and rewrite it as the logarithm of the product and quotient. This is essentially is working backwards from the last objective. Write as the logarithm of a single expression: Ex. 15a

€

23

log(x) + log(y) – 3log(z) Ex. 15b 5ln(x + 1) – 3ln(x2 + 9) + ln(x + 2) Ex. 15c

€

15

log4(x2 + 16) – 2log4(x – 4) – 3log4(x + 3) + 7log4(x) Solution: a)

€

23

log(x) + log(y) – 3log(z) (apply property #5) = log(x2/3) + log(y) – log(z3) (apply property #3) = log(x2/3y) – log(z3) (apply property #4)

= log

€

x2 / 3 yz3

⎛

⎝ ⎜

⎞

⎠ ⎟

= log

€

x23 y

z3

⎛

⎝ ⎜

⎞

⎠ ⎟

b) 5ln(x + 1) – 3ln(x2 + 9) + ln(x + 2) (apply property #5) = ln[(x + 1)5] – ln[(x2 + 9)3] + ln(x + 2) (apply property #3) = ln[(x + 1)5(x + 2)] – ln[(x2 + 9)3] (apply property #4)

= ln

€

(x+1)5(x+2)(x2 + 9)3

⎡

⎣ ⎢ ⎤

⎦ ⎥

176

c)

€

15

log4(x2 + 16) – 2log4(x – 4) – 3log4(x + 3) + 7log4(x) (apply property #5) = log4(

€

x2 +165 ) – log4[(x – 4)2] – log4[(x + 3)3] + log4(x7) (regroup) = log4(

€

x2 +165 ) + log4(x7) – log4[(x – 4)2] – log4[(x + 3)3] (factor out the negative) = log4(

€

x2 +165 ) + log4(x7) – {log4[(x – 4)2] + log4[(x + 3)3]} (apply property #3) = log4(x7

€

x2 +165 ) – log4[(x – 4)2(x + 3)3] (apply property #4)

= log4(

€

x7 x2 +165

(x−4)2(x+3)3)

Caution – Avoid these common errors: loga(u + v) ≠ loga(u) + loga(v) loga(u•v) ≠ loga(u + v) [Correct statement: loga(u•v) = loga(u) + loga(v)] loga(u – v) ≠ loga(u) – loga(v) loga(

€

uv ) ≠

€

loga(u)loga(v)

[Correct statement: loga(

€

uv ) = loga(u) – loga(v)]

[loga(M)]r ≠ rloga(M) loga(Mr) ≠ [loga(M)]r [Correct statement: loga(Mr) = rloga(M)]