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CHAPTER 17: EXPONENTIAL AND LOGARITHM
                    FUNCTIONS


                     1. Exponential Functions
Definition 1.1. An exponential function is a function of the form
                   f (x) = A · bx   (A, b constants)
               t               r
( or f (t) = Ab or f (r) = Ab . . . etc.) That is, it is a function with a
variable exponent.
Example 1.1. The functions f (x) = 2x , g(x) = 3x , h(x) = 100 · 5x are
exponential functions.
Note the difference between the exponential function f (x) = 2x with fixed
base and variable exponent or power, and the power function f (x) = x2
with fixed exponent and variable base.




Exponential functions occur naturally:
    • Radioactive decay
    • unconstrained growth of a population
Example 1.2. The population of Mexico (1980-1983):
                          year pop. (in millions)
                          1980        67.38
                          1981        69.13
                          1982        70.93
                          1983        72.77
The successsive ratios are (approximately) the same:
                               69.13
                                     ≈ 1.026
                               67.38
                               70.93
                                     ≈ 1.026
                               69.13
                                    1
2                      First Science MATH1200 Calculus

In each year, the population increases by a factor of 1.026.
Let P (t) denote the population at time t (where t is the number of years
since 1980):

            P (0) =   67.38 = P0
            P (1) =   67.38 · (1.026)
            P (2) =   67.38 · (1.026) · (1.026) = 67.38 · (1.026)2
            P (3) =   67.38 · (1.026)3
                  .
                  .
                  .
            P (t) =   67.38 · (1.026)t = P0 bt

Thus the estimated population in 1990 is P (10) = 67.38 · (1.026)10 ≈ 87.1




Note: If f (t) = Abt , we get exponential growth if b > 1 and we get expo-
nential decay if b < 1. (b is called the base of the exponential function.)

Example 1.3. Consider the exponential function f (t) = (1/2)t = 2−t :




This function exhibits exponential decay.

Example 1.4. For any given amount of radioactive potassium ( 44 K ),
                                                                 19
the amount remaining one second later is 99.97%. Find a formula for the
amount at time t. (Let A0 be the amount at time 0).
First Science MATH1200 Calculus              3

Solution:
                        A(t) = amount at time t
                        A(0) = A0
                        A(1) = A0 · (0.9997)
                             .
                             .
                             .
                        A(t) = A0 · (0.9997)t

                              2. Half-Lives
Suppose that A(t) = A0 · bt with b < 1
(i.e., we have exponential decay).
Then there is a number h > 0 with
                                        1
                                   bh =
                                        2
Thus, if t is any time, the amount at time t + h is
                 A(t + h) = A0 · bt+h
                           = A0 · b t · b h
                               1
                           =     A0 b t
                               2
                               1
                           =     A(t)
                               2
                               1
                           =      of the amount at time t
                               2
The number h is called the half-life or 1/2-life of the process.
                                44
Example 2.1. The 1/2-life of    19 K   is 22 mins.
Example 2.2. The 1/2-life of Carbon-14 is 5700 years.
A similar computation shows that if we have exponential growth
                           A(t) = A0 bt       b>1
then bd = 2 for some d > 0.
This number d is called the doubling time for the process.
Example 2.3. The 1/2-life of carbon-14 is 5700 years. Find the formula
for the amount at time t (t in years).
Solution: A(t) = A0 bt . What’s b?
We know
                                             1
                         A(5700) = A0 b5700 = A0
                                             2
So
                                         1
                                 b5700 =
                                         2
4                        First Science MATH1200 Calculus
                                     1
                               1   5700
                =⇒ b =                    = (0.5)0.000175 ≈ 0.999878
                               2

So A(t) ≈ A0 · (0.999878)t .

Example 2.4. For a process of exponential decay

                                     A(t) = A0 bt

determine the half-life.
Solution: We must solve
                                                 1
                                          bh =
                                                 2
for h.
To do this, we need logarithms.


                                   3. Logarithms
The logarithm function answers the question: What power of the number a
is equal to the number c?

Definition 3.1. Suppose that a > 0 and ab = c, then we say that b is the
logarithm of c to the base a and write

                                     loga (c) = b

Example 3.1.

                       23 = 8         =⇒         3 = log2 8
                  104 = 10000         =⇒         4 = log10 10000
                           1                                   1
                   10−1 =             =⇒         −1 = log10
                           10                                 10
                         0
                       a =1           =⇒         0 = loga (1)

Thus ‘loga c’ means the power that a must be raised to in order to get c.

Example 3.2. Thus if we ask: What power of 3 is equal to 47? we are
asking for log3 (47). If we ask: what power of 10 is equal to 50?, we are
asking for log10 (50).

Note that the ‘input’ in a logarithm must be a positive number ; i.e., the
domain of the function f (x) = loga (x) is (0, ∞).
Graph of y = log10 x:
First Science MATH1200 Calculus                  5




Notation: In elementary texts (and in this course), log10 x is simply de-
noted log x , and is called the ‘common logarithm’.
Thus log x = y means 10y = x.
Example 3.3. So log(5) = 0.69897 . . . means 100.69897... = 5.

                        4. Properties of Logarithms
Theorem 4.1. Fix a > 0.
  (1) loga (1) = 0
  (2) loga (a) = 1
  (3) loga (c1 · c2 ) = loga (c1 ) + loga (c2 )
              c1
  (4) loga           = loga (c1 ) − loga (c2 )
              c2
  (5) loga (cd ) = d loga (c)
Proof:
  (1) Since a0 = 1.
  (2) Since a1 = a.
  (3) Let b1 = loga (c1 ) and b2 = loga (c2 ). Then ab1 = c1 and ab2 = c2 . So
                         ab1 +b2 = c1 · c2        (First Law )
                 =⇒ loga (c1 c2 ) = b1 + b2 = loga (c1 ) + loga (c2 )
    (4)
                                   c1                   c1
               loga (c1 ) = loga      · c2    = loga          + loga (c2 )
                                   c2                   c2
      (using 3.).
      Now subtract loga (c2 ) from both sides.
  (5) Let b = loga (c). So ab = c.
      Thus abd = (ab )d = cd (Second Law).
      So loga (cd ) = bd = d loga (c)
Note: Taking d = −1 in 5. gives
                                       1
                               loga          = − loga (c)
                                       c
6                       First Science MATH1200 Calculus

Example 4.1. The formula for the amount of radioactive polonium is
                   A(t) = A0 (0.99506)t         (t in days )
What is the half-life?
Solution: We need to solve .99506h = 1/2 for h:
                      log(0.99506h ) = log(1/2)
                    h · log(0.99506) = log(1/2)
                                         log(1/2)
                                   h =
                                       log(0.99506)
                                       ≈ 140 (days )
Thus for a process of exponential decay we have,
                                           log(1/2)
                            half-life =
                                          log(base)
Similarly, for a process of exponential growth,
                                               log(2)
                        doubling time =
                                             log(base)
Example 4.2. Estimate the doubling time of the Mexican population.
Solution:
                          log(2)
                   d=              = 27 (years )
                        log(1.026)
In general, logarithms are useful for solving equations where the unknown
occurs as an exponent:
Example 4.3. Solve 7x = 231 · 5x for x.
Solution: Take logs of both sides:
                           log(7x ) = log(231 · 5x )
                          x log(7) = log(231) + x log(5)
                x(log(7) − log(5)) = log(231)
                                         log 231
                                 x =
                                      log 7 − log 5
                                    ≈ 16.175
         5. Derivatives of logarithms and exponentials
What are
                            d x          d
                              (2 ) or      log(x)?
                           dx           dx
We will not be in a position to answer these questions until we have defined
the natural logarithm function, and to define the natural logarithm we have
to first develop the theory of integration.

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Notes 17

  • 1. CHAPTER 17: EXPONENTIAL AND LOGARITHM FUNCTIONS 1. Exponential Functions Definition 1.1. An exponential function is a function of the form f (x) = A · bx (A, b constants) t r ( or f (t) = Ab or f (r) = Ab . . . etc.) That is, it is a function with a variable exponent. Example 1.1. The functions f (x) = 2x , g(x) = 3x , h(x) = 100 · 5x are exponential functions. Note the difference between the exponential function f (x) = 2x with fixed base and variable exponent or power, and the power function f (x) = x2 with fixed exponent and variable base. Exponential functions occur naturally: • Radioactive decay • unconstrained growth of a population Example 1.2. The population of Mexico (1980-1983): year pop. (in millions) 1980 67.38 1981 69.13 1982 70.93 1983 72.77 The successsive ratios are (approximately) the same: 69.13 ≈ 1.026 67.38 70.93 ≈ 1.026 69.13 1
  • 2. 2 First Science MATH1200 Calculus In each year, the population increases by a factor of 1.026. Let P (t) denote the population at time t (where t is the number of years since 1980): P (0) = 67.38 = P0 P (1) = 67.38 · (1.026) P (2) = 67.38 · (1.026) · (1.026) = 67.38 · (1.026)2 P (3) = 67.38 · (1.026)3 . . . P (t) = 67.38 · (1.026)t = P0 bt Thus the estimated population in 1990 is P (10) = 67.38 · (1.026)10 ≈ 87.1 Note: If f (t) = Abt , we get exponential growth if b > 1 and we get expo- nential decay if b < 1. (b is called the base of the exponential function.) Example 1.3. Consider the exponential function f (t) = (1/2)t = 2−t : This function exhibits exponential decay. Example 1.4. For any given amount of radioactive potassium ( 44 K ), 19 the amount remaining one second later is 99.97%. Find a formula for the amount at time t. (Let A0 be the amount at time 0).
  • 3. First Science MATH1200 Calculus 3 Solution: A(t) = amount at time t A(0) = A0 A(1) = A0 · (0.9997) . . . A(t) = A0 · (0.9997)t 2. Half-Lives Suppose that A(t) = A0 · bt with b < 1 (i.e., we have exponential decay). Then there is a number h > 0 with 1 bh = 2 Thus, if t is any time, the amount at time t + h is A(t + h) = A0 · bt+h = A0 · b t · b h 1 = A0 b t 2 1 = A(t) 2 1 = of the amount at time t 2 The number h is called the half-life or 1/2-life of the process. 44 Example 2.1. The 1/2-life of 19 K is 22 mins. Example 2.2. The 1/2-life of Carbon-14 is 5700 years. A similar computation shows that if we have exponential growth A(t) = A0 bt b>1 then bd = 2 for some d > 0. This number d is called the doubling time for the process. Example 2.3. The 1/2-life of carbon-14 is 5700 years. Find the formula for the amount at time t (t in years). Solution: A(t) = A0 bt . What’s b? We know 1 A(5700) = A0 b5700 = A0 2 So 1 b5700 = 2
  • 4. 4 First Science MATH1200 Calculus 1 1 5700 =⇒ b = = (0.5)0.000175 ≈ 0.999878 2 So A(t) ≈ A0 · (0.999878)t . Example 2.4. For a process of exponential decay A(t) = A0 bt determine the half-life. Solution: We must solve 1 bh = 2 for h. To do this, we need logarithms. 3. Logarithms The logarithm function answers the question: What power of the number a is equal to the number c? Definition 3.1. Suppose that a > 0 and ab = c, then we say that b is the logarithm of c to the base a and write loga (c) = b Example 3.1. 23 = 8 =⇒ 3 = log2 8 104 = 10000 =⇒ 4 = log10 10000 1 1 10−1 = =⇒ −1 = log10 10 10 0 a =1 =⇒ 0 = loga (1) Thus ‘loga c’ means the power that a must be raised to in order to get c. Example 3.2. Thus if we ask: What power of 3 is equal to 47? we are asking for log3 (47). If we ask: what power of 10 is equal to 50?, we are asking for log10 (50). Note that the ‘input’ in a logarithm must be a positive number ; i.e., the domain of the function f (x) = loga (x) is (0, ∞). Graph of y = log10 x:
  • 5. First Science MATH1200 Calculus 5 Notation: In elementary texts (and in this course), log10 x is simply de- noted log x , and is called the ‘common logarithm’. Thus log x = y means 10y = x. Example 3.3. So log(5) = 0.69897 . . . means 100.69897... = 5. 4. Properties of Logarithms Theorem 4.1. Fix a > 0. (1) loga (1) = 0 (2) loga (a) = 1 (3) loga (c1 · c2 ) = loga (c1 ) + loga (c2 ) c1 (4) loga = loga (c1 ) − loga (c2 ) c2 (5) loga (cd ) = d loga (c) Proof: (1) Since a0 = 1. (2) Since a1 = a. (3) Let b1 = loga (c1 ) and b2 = loga (c2 ). Then ab1 = c1 and ab2 = c2 . So ab1 +b2 = c1 · c2 (First Law ) =⇒ loga (c1 c2 ) = b1 + b2 = loga (c1 ) + loga (c2 ) (4) c1 c1 loga (c1 ) = loga · c2 = loga + loga (c2 ) c2 c2 (using 3.). Now subtract loga (c2 ) from both sides. (5) Let b = loga (c). So ab = c. Thus abd = (ab )d = cd (Second Law). So loga (cd ) = bd = d loga (c) Note: Taking d = −1 in 5. gives 1 loga = − loga (c) c
  • 6. 6 First Science MATH1200 Calculus Example 4.1. The formula for the amount of radioactive polonium is A(t) = A0 (0.99506)t (t in days ) What is the half-life? Solution: We need to solve .99506h = 1/2 for h: log(0.99506h ) = log(1/2) h · log(0.99506) = log(1/2) log(1/2) h = log(0.99506) ≈ 140 (days ) Thus for a process of exponential decay we have, log(1/2) half-life = log(base) Similarly, for a process of exponential growth, log(2) doubling time = log(base) Example 4.2. Estimate the doubling time of the Mexican population. Solution: log(2) d= = 27 (years ) log(1.026) In general, logarithms are useful for solving equations where the unknown occurs as an exponent: Example 4.3. Solve 7x = 231 · 5x for x. Solution: Take logs of both sides: log(7x ) = log(231 · 5x ) x log(7) = log(231) + x log(5) x(log(7) − log(5)) = log(231) log 231 x = log 7 − log 5 ≈ 16.175 5. Derivatives of logarithms and exponentials What are d x d (2 ) or log(x)? dx dx We will not be in a position to answer these questions until we have defined the natural logarithm function, and to define the natural logarithm we have to first develop the theory of integration.