The document discusses properties and laws of exponents, radicals, logarithms including the definition of rational exponents, properties of logarithms such as the change of base formula, and examples of simplifying expressions using exponent laws, combining like radicals, and solving logarithmic equations by using properties of logarithms. It also provides sample problems and their step-by-step solutions for simplifying expressions and solving equations involving exponents, radicals, and logarithms.

1. Chapter 3
Radical and
exponents

2. Exponential notation
 represent as to the th power .
Exponent
Base (integers)
(real number)

3. General case Special cases
(n is any positive integers)
Zero and negative exponent Example
(where a c ≠ 0)

4. Law of exponents
Law Example

5. Theorem on negative exponents
Prove:
Prove:

6. Example :
simplifying negative exponents
(1) 1 4 3 2
( x y )
3
1 2 4 2 3 2
( ) (x ) ( y )
3
2 8 6
3 x y
6
y
9 8
x

7. Principal nth root
Where n=positive integer greater than 1
= real number
Value for Value for
= positive real number b
Such that
=negative real number b
Such that

8. Properties of
RADICAL
index
radicand
Radical sign
PROPERTY EXAMPLE

9. Example:
combining radicals
Question: 4
α
3
α2
1
α 4
2
3
α
( 1 2 ) 5
α 4 3
α 12
1
5
α 12
1
12
α5

10. Law of radicals
law example
WARNING!

11. Example:
Removing factors from radicals
Question:
3a 2 b 3 6 a 5b
3a 2 b 3 .2.3a 5b
(32 a 6 b 4 )(2a )
3 2 2
(3a b ) ( 2a )
(3a 3b 2 ) 2 2a
3a 3b 2 2a

12. Rationalizing a denominator
Factor in denominator Multiply numerator Resulting factor
and denominator by

13. How do you know when a
radical problem is done?
(1) No radicals can be simplified.
Example:
8
(2) There are no fractions in the radical.
Example: 1
4
(3) There are no radicals in the denominator.
Example: 1
5

14. Example :
Rationalizing denominators
(1)
=
(2)
=

15. Definition of rational
exponents
m/n = rational number
n = positive integer greater than 1
= real number, then
(1)
(2)
(3)

16. Example:
Simplifying rational powers
(1)
6 4
x y
6 4
2 2
x .y
3 2
x y

17. How do you simplify variables in the radical?
7
x
Look at these examples and try to find the pattern…
1
x x What is the answer to x7 ?
2
x x x 7
x 3
x
3
x x x
4 2
As a general rule, divide the
x x exponent by two. The
x 5 2
x x remainder stays in the
6 3 radical.
x x

18. LOGARITHMS

19. Definition of
• The logarithms of with base is defined by:
exponent
if and only if
base
For every and every real number .

20. Illustration
Logarithmic form Exponential form

21. • The logarithmic function with base is one-to-
one. Thus, the following equivalent conditions
are satisfied for positive real number x1 and x2 .
(1) If x1 x2 , then .
(2) If , then x1 x2 .

22. Example :
Solving a logarithms equation.
Check..
Since is a true statement, then

23. • Definition of common logarithm:
for every
• Defition of natural logarithm:
for every

24. Properties of logarithms
Logarithms with base Common logarithms Natural logarithms

25. = Power to which you need to raise 2 in order to get 8
(a) log28
= 3 ( Since 23 = 8 )
= Power to which you need to raise 4 in order to get 1
(b) log41
= 0 ( Since 40 = 1 )
= Power to which you need to raise 10 in order to get 10,000
(c) log1010,000
= 4 ( Since 104 = 10,000 )
= Power to which you need to raise 10 in order to get 1/100
(d) log101/100
= 2 ( Since 10-2 = 1/100 )

26. Laws of logarithms
Common logarithms Natural logarithms

27. Example:
Application law of logarithm
• log abc²
d3
= log (abc²) − log d 3
= log a + log b + log c² − log d 3
= log a + log b + 2 log c − 3 log d

28. Change of base formula
• If and if and are positive real
number, then

29. Special change of base formula

30. Example

31. Example:
Solve
Solution :

32. QUESTION

33. Question 1
Simplify:

34. Question 2

35. Question 3
Simplifying:

36. Question 4
Simplifying:

37. Question 5:

38. Question 6
• Solve logb(x2) = logb(2x – 1).
x2 = 2x – 1
x2 – 2x + 1 = 0
(x – 1)(x – 1) = 0
Then the solution is x = 1.

39. Question 7
• Solve ln( ex ) = ln( e3 ) + ln( e5 )
ln( ex ) = ln( e3 ) + ln( e5 )
ln( ex ) = ln(( e3 )( e5 ))
ln( ex ) = ln( e3 + 5 )
ln( ex ) = ln( e8 )
Comparing the arguments:
ex = e8
x=8

40. Question 8
Solve log2(x) + log2(x – 2) = 3
log2(x) + log2(x – 2) = 3
log2((x)(x – 2)) = 3
log2(x2 – 2x) = 3
23 = x2 – 2x
8 = x2 – 2x
0 = x2 – 2x – 8
0 = (x – 4)(x + 2)
x = 4, –2
Since logs cannot have zero or negative arguments, then the
solution to the original equation cannot be x = –2.
Solution: x=4

41. Question 9
• SOLUTION:

42. Therefore or

43. Question 10
If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1,
SOLUTION :
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10
log10 [5 (5x + 1)] = log10 [10(x + 5)]
5(5x + 1) = 10(x + 5)
5x + 1 = 2x + 10
3x = 9
x=3

44. Thank you…
Prepared by:
Nurul Atiyah binti Ripin
(D20111048011)
Irma Naziela binti Rosli
(D20111048007)