this is the topi on radical and exponents in

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)