1. Copyright © Cengage Learning. All rights reserved.
5. Inverse, Exponential and
Logarithmic Functions
5.5 Properties of Logarithms
5.6 Exponential and Logarithmic Equations
Copyright © Cengage Learning. All rights reserved.

2. Properties of Logarithms
Logarithm of x, that is loga x, can be interpreted as an exponent. Thus, it seems reasonable to expect that the laws of exponents can be used to obtain corresponding laws of logarithms. This is demonstrated in the proofs of the following laws, which are fundamental for all work with logarithms.

3. Properties of Logarithms
The laws of logarithms for the special cases a = 10 (common logs) and a = e (natural logs) are written as shown in the following chart.

4. Properties of Logarithms
As indicated by the following warning, there are no laws for expressing loga (u + w) or loga (u – w) in terms of simpler logarithms.
The next example illustrates the uses of the laws of logarithms.

5. Example 1 – Using laws of logarithms
Express in terms of logarithms of x, y, and z.
Solution:
We write as y1/2 and use laws of logarithms:
= loga (x3y1/2) – loga z2
= loga x3 + loga y1/2 – loga z2
= 3 loga x + loga y – 2 loga z
law 2
law 1
law 3

6. Example 1 – Solution
cont’d
Note that if a term with a positive exponent (such as x3) is in the numerator of the original expression, it will have a positive coefficient in the expanded form, and if it is in the denominator (such as z2), it will have a negative coefficient in the expanded form.

7. Example 3 – Solving a logarithmic equation
Solve the equation log5 (2x + 3) = log log5 3.
Solution:
log5 (2x + 3) = log log5 3
log5 (2x + 3) = log5 (11  3)
2x + 3 = 33
x = 15
given
law 1 of logarithms
logarithmic functions are one-to-one
solve for x

8. Example 3 – Solution Check: x = 15 LS: log5 (2  15 + 3) = log5 33
cont’d
Check: x = 15 LS: log5 (2  ) = log5 33
RS: log log5 3 = log5 (11  3) = log5 33
Since log5 33 = log5 33 is a true statement, x = 15 is a solution.

9. Example 6 – Shifting the graph of a logarithmic equation
Sketch the graph of y = log3 (81x).
Solution:
We may rewrite the equation as follows:
y = log3 (81x)
= log log3 x
= log log3 x
given
law 1 of logarithms
81 = 34

10. Example 6 – Solution = 4 + log3 x
cont’d
= 4 + log3 x
Thus, we can obtain the graph of y = log3 (81x) by vertically shifting the graph of y = log3 x upward four units.
This gives us the sketch in Figure 3.
loga ax = x
Figure 3

11. Example 7 – Sketching graphs of logarithmic equations
Sketch the graph of the equation:
(a) y = log3 (x2) (b) y = 2 log3 x
Solution:
(a) Since x2 = | x |2, we may rewrite the given equation as
y = log3 | x |2.
Using law 3 of logarithms, we have
y = 2 log3 | x |.

12. Example 7 – Solution
cont’d
We can obtain the graph of y = 2 log3 | x | by multiplying the y-coordinates of points on the graph of y = log3 | x | by 2.
This gives us the graph in Figure 4(a).
Figure 4(a)

13. Example 7 – Solution
cont’d
(b) If y = 2 log3 x, then x must be positive. Hence, the graph is identical to that part of the graph of y = 2 log3 | x | in Figure 4(a) that lies to the right of the y-axis.
This gives us Figure 4(b).
Figure 4(b)

14. Example 8 – A relationship between selling price and demand
In the study of economics, the demand D for a product is often related to its selling price p by an equation of the form
loga D = loga c – k loga p,
where a, c, and k are positive constants.
(a) Solve the equation for D.
(b) How does increasing or decreasing the selling price affect the demand?

15. Example 8 – Solution (a) loga D = loga c – k loga p
loga D = loga c – loga pk
loga D =
D =
given
law 3 of logarithms
law 2 of logarithms
loga is one-to-one

16. Example 8 – Solution
cont’d
(b) If the price p is increased, the denominator pk in D = c/pk will also increase and hence the demand D for the product will decrease.
If the price is decreased, then pk will decrease and the demand D will increase.

17. Exponential and Logarithmic Equations
In this section we shall consider various types of exponential and logarithmic equations and their applications.
When solving an equation involving exponential expressions with constant bases and variables appearing in the exponent(s), we often equate the logarithms of both sides of the equation.
When we do so, the variables in the exponent become multipliers, and the resulting equation is usually easier to solve. We will refer to this step as simply “take log of both sides.”

18. Example 1 – Solving an exponential equation
Solve the equation 3x = 21.
Solution:
3x = 21
log (3x) = log 21
x log 3 = log 21
given
take log of both sides
law 3 of logarithms
divide by log 3

19. Example 1 – Solution
cont’d
We could also have used natural logarithms to obtain
Using a calculator gives us the approximate solution x  2.77.
A partial check is to note that since 32 = 9 and 33 = 27, the number x such that 3x = 21 must be between 2 and 3, somewhat closer to 3 than to 2.

20. Exponential and Logarithmic Equations
We could also have solved the equation in Example 1 by changing the exponential form 3x = 21 to logarithmic form, Obtaining
x = log3 21.
This is, in fact, the solution of the equation; however, since calculators typically have keys only for log and ln, we cannot approximate log3 21 directly.
The next theorem gives us a simple change of base formula for finding logb u if u > 0 and b is any logarithmic base.

21. Exponential and Logarithmic Equations
The following special case of the change of base formula is obtained by letting u = a and using the fact that loga a = 1:
The change of base formula is sometimes confused with law 2 of logarithms.

22. Exponential and Logarithmic Equations
The first of the following warnings could be remembered with the phrase “a quotient of logs is not the log of the quotient.”

23. Exponential and Logarithmic Equations
The most frequently used special cases of the change of base formula are those for a = 10 (common logarithms) and a = e (natural logarithms), as stated in the following box.

24. Example 2 – Using a change of base formula
Solve the equation 3x = 21.
Solution:
We proceed as follows:
3x = 21
x = log3 21
given
change to logarithmic form
special change of base formula 1

25. Example 2 – Solution
cont’d
Another method is to use special change of base formula 2, obtaining

26. Exponential and Logarithmic Equations
Logarithms with base 2 are used in computer science.
The next example indicates how to approximate logarithms with base 2 using change of base formulas.

27. Example 3 – Approximating a logarithm with base 2
Approximate log2 5 using
(a) common logarithms (b) natural logarithms
Solution:
Using special change of base formulas 1 and 2, we obtain the following:
 2.322

28. Example 6 – Solving an equation involving logarithms
Solve the equation log for x.
Solution:
log x1/3 =
log x =
(log x)2 = log x
(log x)2 = 9 log x
= x1/n
log xr = r log x
square both sides
multiply by 9

29. Example 6 – Solution (log x)2 – 9 log x = 0 (log x)(log x – 9) = 0
cont’d
(log x)2 – 9 log x = 0
(log x)(log x – 9) = 0
log x = 0, log x – 9 = 0
log x = 9
x = 100 = 1 or x = 109
make one side 0
factor out log x
set each factor equal to 0
add 9
log10 x = a x = 10a

30. Example 6 – Solution Check: x = 1 LS: log = log 1 = 0 RS:
cont’d
Check: x = LS: log = log 1 = 0
RS:
Check: x = LS: log = log 103 = 3
The equation has two solutions, 1 and 1 billion.

31. Exponential and Logarithmic Equations
The function y = 2/(ex + e–x) is called the hyperbolic secant function. In the next example we solve this equation for x in terms of y.
Under suitable restrictions, this gives us the inverse function.

32. Example 7 – Finding an inverse hyperbolic function
Solve y = 2/(ex + e–x) for x in terms of y.
Solution:
yex + ye–x = 2
yex = 2
yex(ex) = 2(ex)
y(ex)2 – 2ex + y = 0
given
multiply by ex + e–x
definition of negative exponent
multiply by the lcd, ex
simplify and subtract 2ex

33. Example 7 – Solution
cont’d
We recognize this form of the equation as a quadratic in ex with coefficients a = y, b = –2, and c = y. Note that we are solving for ex, not x.
quadratic formula
simplify
factor out

34. Example 7 – Solution
cont’d
For the blue curve y = f (x) in Figure 2, the inverse function is
y = f –1(x) = ln
shown in blue in Figure 3.
cancel a factor of 2
take ln of both sides
Figure 2

35. Example 7 – Solution
cont’d
Notice the domain and range relationships. For the red curve y = g (x) in Figure 2, the inverse function is
shown in red in Figure 3.
Figure 3

36. Example 7 – Solution
cont’d
Since the hyperbolic secant is not one-to-one, it cannot have one simple equation for its inverse.

37. Exponential and Logarithmic Equations
The inverse hyperbolic secant is part of the equation of the curve called a tractrix.

38. Example 10 – A logistic curve
A logistic curve is the graph of an equation of the form
where k, b, and c are positive constants. Such curves are useful for describing a population y that grows rapidly initially, but whose growth rate decreases after x reaches a certain value.
In a famous study of the growth of protozoa by Gause, a population of Paramecium caudata was found to be described by a logistic equation with c = , k = 105 and x the time in days.

39. Example 10 – A logistic curve
cont’d
(a) Find b if the initial population was 3 protozoa.
(b) In the study, the maximum growth rate took place at y = 52. At what time x did this occur?
(c) Show that after a long period of time, the population described by any logistic curve approaches the constant k.

40. Example 10 – Solution
(a) Letting c = and k = 105 in the logistic equation, we obtain
We now proceed as follows:
1 + b = 35
b = 34
let y = 3 when x = 0
multiply by
solve for b

41. Example 10 – Solution
cont’d
(b) Using the fact that b = 34 leads to the following:
1 + 34e–1.1244x =
let y = 52 in part (a)
multiply by
isolate e–1.1244x
change to logarithmic form
divide by –1.1244

42. Example 10 – Solution
cont’d
(c) As x  , e–cx  0. Hence,