1. and Logarithmic Functions
Review of Exponential
and Logarithmic Functions
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

2. Exponential Functions
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

3. An exponential function is a function of the form
where a is a positive real number (a > 0) and The domain of f is the set of all real numbers.
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

4. (1, 6)
(1, 3)
(-1, 1/6)
(-1, 1/3)
(0, 1)
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

5. a >1 Summary of the characteristics of the graph of
The domain is all real numbers. Range is set of positive numbers.
No x-intercepts; y-intercept is 1.
The x-axis (y=0) is a horizontal asymptote as
With a>1, is an increasing function and is one-to-one.
The graph contains the points (0,1); (1,a), and (-1, 1/a).
The graph is smooth continuous with no corners or gaps.
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

6. (-1, 6)
(-1, 3)
(0, 1)
(1, 1/3)
(1, 1/6)
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

7. 0 <a <1 Summary of the characteristics of the graph of
The domain is all real numbers. Range is set of positive numbers.
No x-intercepts; y-intercept is 1.
The x-axis (y=0) is a horizontal asymptote as
With 0<a<1, is a decreasing function and is one-to-one.
The graph contains the points (0,1); (1,a), and (-1, 1/a).
The graph is smooth continuous with no corners or gaps.
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

8. Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

9. (1, 3) (0, 1) Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

10. (-1, 3) (0, 1) Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

11. Domain: All real numbers Range: { y | y >2 } or
(-1, 5)
(0, 3)
y = 2
Domain: All real numbers
Range: { y | y >2 } or
Horizontal Asymptote: y = 2
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

12. More Exponential Functions (Shifts) :
An equation in the form f(x) = ax.
Recall that if 0 < a < 1 , the graph represents exponential decay
and that if a > 1, the graph represents exponential growth
Examples: f(x) = (1/2)x f(x) = 2x
Exponential Decay
Exponential Growth
We will take a look at how these graphs “shift” according to changes in their equation...
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

13. f(x) = (1/2)x f(x) = (1/2)x + 1 f(x) = (1/2)x – 3
Take a look at how the following graphs compare to the original graph of f(x) = (1/2)x :
f(x) = (1/2)x f(x) = (1/2)x f(x) = (1/2)x – 3
Vertical Shift: The graphs of f(x) = ax + k are shifted vertically by k units.
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

14. f(x) = (2)x f(x) = (2)x – 3 f(x) = (2)x + 2 – 3
Take a look at how the following graphs compare to the original graph of f(x) = (2)x :
f(x) = (2)x f(x) = (2)x – f(x) = (2)x + 2 – 3
(3,1)
(0,1)
(-2,-2)
Notice that f(0) = 1
Notice that this graph
is shifted 3 units to the
right.
Notice that this graph
is shifted 2 units to the
left and 3 units down.
Horizontal Shift: The graphs of f(x) = ax – h are shifted horizontally by h units.
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

15. f(x) = (2)x f(x) = –(2)x f(x) = –(2)x + 2 – 3
Take a look at how the following graphs compare to the original graph of f(x) = (2)x :
f(x) = (2)x f(x) = –(2)x f(x) = –(2)x + 2 – 3
(0,1)
(0,-1)
(-2,-4)
Notice that f(0) = 1
This graph
is a reflection of
f(x) = (2)x . The graph is
reflected over the x-axis.
Shift the graph of
f(x) = (2)x ,2 units to the left. Reflect the graph over the x-axis. Then, shift the graph 3 units down
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

16. Logarithmic Functions
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

17. Logarithmic Functions
A logarithmic function is the inverse of an exponential function.
For the function y = 2x, the inverse is x = 2y.
In order to solve this inverse equation for y,
we write it in logarithmic form.
x = 2y is written as y = log2x
and is read as “y = the logarithm of x to base 2”.
y = 2x 1 2 4 8 16 1 2 4 8 16
y = log2x
(x = 2y)
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

18. Graphing the Logarithmic Function
y = x
y = 2x
y = log2x
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

19. Comparing Exponential and Logarithmic Function Graphs
y = 2x
y = log2x
The y-intercept is 1.
There is no y-intercept.
There is no x-intercept.
The x-intercept is 1.
The domain is All Reals
The domain is x > 0
The range is y > 0.
The range is All Reals
There is a horizontal asymptote
at y = 0.
There is a vertical asymptote
at x = 0.
The graph of y = 2x has been reflected in the line of y = x,
to give the graph of y = log2x.
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

20. 2 is the exponent of the power, to which 7 is raised, to equal 49.
Logarithms
Consider 72 = 49.
2 is the exponent of the power, to which 7 is raised, to equal 49.
The logarithm of 49 to the base 7 is equal to 2
(log749 = 2).
Exponential
notation
Logarithmic
form
log749 = 2
72 = 49
In general: If bx = N,
then logbN = x.
State in logarithmic form:
State in exponential form:
a) 63 = 216
log6216 = 3
a) log5125 = 3
53 = 125
b) log2128= 7
27 = 128
b) 42 = 16
log416 = 2
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

21. State in logarithmic form:
Logarithms
State in logarithmic form: a) b)
log2 32 = 3x + 2
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

22. Evaluating Logarithms
Think – What power must you raise 2 to, to get 128?
2. log327
Note:
log2128 = log227
= 7
log327 = log333
= 3
log2128 = x
2x = 128
2x = 27
x = 7
log327 = x
3x = 27
3x = 33
x = 3
logaam = m because logarithmic and
exponential functions are
inverses.
3. log556
= 6
4. log816
5. log81
log816 = x
8x = 16
23x = 24
3x = 4
log81 = x
8x = 1
8x = 80
x = 0
loga1 = 0
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

23. Evaluating Logarithms6.
log4(log338) 7.
= x
log48 = x
4x = 8
22x = 23
2x = 3
2x = 1
9. Given log165 = x, and log84 = y,
express log220 in terms of x and y. 8.
log165 = x
log84 = y
= 23
= 8
16x = 5
24x = 5
8y = 4
23y = 4
log220 = log2(4 x 5)
= log2(23y x 24x)
= log2(23y + 4x)
= 3y + 4x
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

24. Logarithmic Functions
x = 2y is an exponential equation.
Its inverse (solving for y) is called a logarithmic equation.
Let’s look at the parts of each type of equation:
Exponential Equation x = ay
Logarithmic Equation
y = loga x
exponent
/logarithm
base
number
It is helpful to remember: “The logarithm of a number is the power to which the base must be raised to get the given number.”
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

25. Example: Rewrite in exponential form and solve loga64 = 2
base
number
exponent
a2 = 64
a = 8
Example: Solve log5 x = 3
Rewrite in exponential form:
53 = x
x = 125
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

26. Example: Solve
7y =
y = –2
An equation in the form y = logb x where b > 0 and b ≠ 1 is called a logarithmic function.
Logarithmic and exponential functions are inverses of each other
logb bx = x
blogb x = x
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

27. Examples. Evaluate each: a. log8 84 b. 6[log6 (3y – 1)]
logb bx = x
log8 84 = 4
blogb x = x
6[log6 (3y – 1)] = 3y – 1
Here are some special logarithm values:
1. loga 1 = because a0 = 1
2. loga a = because a1 = a
3. loga ax = x because ax = ax
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

28. Laws of
Logarithms
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

29. Laws of Logarithms Consider the following two problems:
Simplify log3 (9 • 27)
= log3 (32• 33)
= log3 (32 + 3) =
Simplify log3 9 + log3 27
= log log3 33 =
These examples suggest the Law:
Product Law of Logarithms:
For all positive numbers m, n and b where b ≠ 1, logb mn = logb m + logb n
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

30. Consider the following: a. b.
= log
= log3 34 – 3
= – 3
= log3 34 – log3 33
= – 3
These examples suggest the following Law:
Quotient Law of Logarithms:
For all positive numbers m, n and b where b ≠ 1, logb m = logb m – logb n n
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

31. The logarithm of a product equals the sum of the logarithms.
The Product and Quotient Laws
Product Law:
logb(mn) = logbm + logbn
The logarithm of a product equals
the sum of the logarithms.
Quotient Law:
The logarithm of a quotient equals
the difference of the logarithms.
Express
as a sum and difference of logarithms:
= log3A + log3B - log3C
Evaluate: log210 + log212.8
= log2(10 x 12.8)
= log2(128)
= log2(27)
= 7
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

32. Simplifying Logarithms
Solve: x = log550 - log510
x = log55
= 1
Given that log79 = 1.129, find the value of log763:
log763 = log7(9 x 7)
= log79 + log77 =
= 2.129
Evaluate: x = log45a + log48a3 - log410a4
x = 1
x = log44
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

33. Consider the following: Evaluate a. log3 94 b. 4 log3 9
= • 4
= (log3 32) • 4
= • 4
These examples suggest the following Law:
Power Law of Logarithms:
For all positive numbers m, n and b where b ≠ 1, logb mp = p • logb m
Consider the following: Evaluate a. log b. 4 log3 9
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

34. The logarithm of a number to a power equals the
The Power Law
Power Law:
logbmn = n logbm
The logarithm of a number to a power equals the
power times the logarithm of the number.
Express as a single log:
= log5216
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

35. Applying the Power Laws
Evaluate:
= 2(1)
+ 4(1)
Given that log62 = and log65 = solve
Thursday, November 29, 2018Thursday, November 29, 2018
= 0.418
Mr M Kennedy

36. Applying the Power Laws
Evaluate:
If log28 = x, express each in terms of x:
log28 = x
log223 = x
3log22 = x
a) log2512
b) log22
= log283
= 3log28
= 3x
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

37. = log
= log12 9 – log12 12
= – 1
= –0.116
= log
= log12 18 – log12 9
= – 0.884 =
More examples: Given log12 9 = and log12 18 = 1.163, find each: a b. log12 2
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

38. Example: Solve 2 log6 4 – 1 log6 8 = log6 x 3
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

39. Natural Exponential Functions
The most commonly used base for exponential and logarithmic functions is the irrational number e.
Exponential functions to base e are called natural exponential functions and are written y = ex.
Natural exponential function follows the same rules as other exponential functions.
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

40. Exponential Function y > 0 for all x passes through (0,1)
positive slope increasing
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

41. eln x = x [Let’s y = ln x and x = ey  x = elnx]
Natural Logarithms
logarithms to base e ( )
loge x or ln x (Note: These mean the same thing!)
ln x is simply the exponent or power to which e must be raised to get x.
y = ln x  x = ey
Since natural exponential functions and natural logarithmic functions are inverses of each other, one is generally helpful in solving the other. Mindful that ln x signifies the power to which e must be raised to get x, for a > 0,
eln x = x [Let’s y = ln x and x = ey  x = elnx]
ln ex = x [Let’s y = ln ex  ey = ex  y = x]
eln x = ln ex = x
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

42. ln e = ln 1 = ln 2 = ln 40 = ln 0.1 = Ex) the natural logarithm of x
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

43. ln e = 1 since e1 = e ln 1 = 0 since e0 = 1
Ex) the natural logarithm of x
ln e = since e1 = e
ln 1 = since e0 = 1
ln 2 = since e = 2
ln 40 = since e = 40
ln 0.1 = since e = -1
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy

44. Natural Logarithmic Function
y > 0 for x > 1
y < 0 for 0 < x < 1
passes through (1,0)
positive slope (increasing)
Thursday, November 29, 2018Thursday, November 29, 2018
Mr M Kennedy