Radical and exponents

Chapter 3

Exponential notation

represent as to the th power .

Exponent (integers)Base

(real number)

General case (n is any positive integers)

Special cases

Zero and negative exponent(where a c ≠ 0)

Example

Law of exponentsLaw Example

Theorem on negative exponents

Prove:

Prove:

Example :simplifying negative exponents

(1)

8

6

682

23242

234

9

3

)()()3

1(

)3

1(

x

y

yx

yx

yx

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

Properties of RADICAL

radicandindex

Radical sign

PROPERTY EXAMPLE

Example:combining radicals

12 5

125

125

32

41

41

3 2

4

1

1

32

α

α

αα

α

α

α

α

)(

Question:

Law of radicals

law example

WARNING!

Example:Removing factors from radicals

Question:

aba

aba

aba

aba

baba

baba

23

2)3(

)2()3(

)2)(3(

3.2.3

63

23

223

223

462

532

532

Rationalizing a denominator

Factor in denominator Multiply numerator and denominator by

Resulting factor

How do you know when a radical problem is done?

(1) No radicals can be simplified.Example:

(2) There are no fractions in the radical.Example:

(3) There are no radicals in the denominator.Example:

8

1

4

1

5

Example :Rationalizing denominators

(1)

=

(2)

=

Definition of rational exponents

m/n = rational number n = positive integer greater than 1 = real number, then

(1)

(2)

(3)

Example: Simplifying rational powers

(1)

23

2

4

2

6

46

.

yx

yx

yx

Look at these examples and try to find the pattern…

How do you simplify variables in the radical?

x7

1x x2x x3x x x4 2x x5 2x x x6 3x x

What is the answer to ? x7

7 3x x x

As a general rule, divide the exponent by two. The remainder stays in the

radical.

LOGARITHMS

Definition of

• The logarithms of with base is defined by:

if and only if

For every and every real number .

base

exponent

Illustration

Logarithmic form Exponential form

• The logarithmic function with base is one-to-one. Thus, the following equivalent conditions are satisfied for positive real number and .

(1) If , then . (2) If , then .

1x 2x

21 xx 21 xx

Example :Solving a logarithms equation.

Since is a true statement, then

Check..

• Definition of common logarithm: for every

• Defition of natural logarithm: for every

Properties of logarithmsLogarithms with base Common logarithms Natural logarithms

(a) log28=

=

Power to which you need to raise 2 in order to get 8

3 ( Since 23 = 8 )

(b) log41=

=

Power to which you need to raise 4 in order to get 1

0 ( Since 40 = 1 )

(c) log1010,000=

=

Power to which you need to raise 10 in order to get 10,000

4 ( Since 104 = 10,000 )

(d) log101/100=

=

Power to which you need to raise 10 in order to get 1/100

2 ( Since 10-2 = 1/100 )

Laws of logarithmsCommon logarithms Natural logarithms

WARNING!!

Example:Application law of logarithm

• log abc² d 3

= log (abc²) − log d 3 = log a + log b + log c² − log d 3 = log a + log b + 2 log c − 3 log d

Change of base formula

• If and if and are positive real number, then

WARNING!!

Special change of base formula

Example

Example:

Solve Solution :

QUESTION

Question 1

Simplify:

Question 2

Question 3Simplifying:

Question 4Simplifying:

Question 5:

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.

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

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

Question 9

• SOLUTION:

Therefore or

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

Thank you…Prepared by:

Nurul Atiyah binti Ripin (D20111048011)

Irma Naziela binti Rosli (D20111048007)