Surds, Indices, and Logarithms Radical

Surds, Indices, and Logarithms

Radical

Definition of the Radical

For all real , x y>0, and all integers a>0, a

x= y if and only if x=ya

where a is the index is the radical

x is the radicand.

Surds

A number which can be expressed as a fraction of integers (assuming the denominator is never 0) is called a rational number. Examples of rational numbers are 5

2, 4 5

− and 2.

A number which cannot be expressed as a fraction of two integers is called an irrational number. Examples of irrational numbers are 2 , 37 and π.

An irrational number involving a root is called a surd. Surds occur frequently in trigonometry, calculus and coordinate geometry. Usually, the exact value of a surd cannot be determined but an approximate value of it can be found by using calculators or mathematical tables. In this chapter,

a means the positive square root of a while − a means the negative square root of a.

General Rules of Surds

Multiplication of surds

a× b= a b×

Division of surds

a a

a b

b b

÷ = =

For example (i) 72 2 72 36 6

2

÷ = = =

(ii) 45 5 45 9 3

5

÷ = = =

These rules are useful for simplifying two or more surds of for combining them into one single surd.

Note, however, that 3+ 6≠ 3 6+ and 6− 2≠ 6 2− which can be easily checked by a calculator; and, therefore, in general a+ b≠ a b+ and a− b ≠ a b− .

Example 1

(i) 3 5+ 5= +(3 1) 5 (ii) 40= 4 10× = 4× 10

=4 5 =2 10

Example 2

Simplify (i) 243− 12+2 75 (ii) 50+ 8+ 32

Solution:

(i) 243− 12+2 75 = 81 3× − 4 3× +2 25 3× = 81× 3− 4× 3+2 25× 3 =9 3− ×2 3 10+ × 3

= − +(9 2 10) 3 17 3=

(ii) 50+ 8+ 32= 25 2× + 4 2× + 16 2× = 25× 2+ 4× 2+ 16× 2 = ×5 2+ ×2 2+ ×4 2

= + +(5 2 4) 2 11 2=

Try This 1

Rationalization of the Denominator

When a fraction has a surd in its denominator, e.g. 3

2, it is usual to eliminate the surd in the denominator. In fact, the writing of surds in the denominators of fractions should be avoided. The process of removing this surd is called rationalizing of the denominator.

m+ n and m− n are specially related surds known as conjugate surds. The product of conjugate surds is always a rational number.

2 2

( m+ n)( m− n)=( m) −( n) = −m n

For example ( 9+ 5)( 9− 5)=( 9)2−( 5)2= − =9 5 4

2 2

( 7+ 3)( 7− 3)=( 7 ) −( 3) = − =7 3 4

Example 3 Example 4

Simplify 5

3. Simplify

4 7+ 3.

Solution: Solution:

5 5 3

3 = 3× 3

4 4 7 3

7 3 7 3 7 3

−

= ×

+ + −

5 3 3

= 4( 7 3)

7 3 − =

− = 7− 3 Example 5

Simplify, without using tables or calculators, the value of 1 1 3− 2+3+ 2.

Solution:

1 1 (3 2) (3 2)

3 2 3 2 (3 2)(3 2) (3 2)(3 2)

+ −

+ = +

− + − + + −

(3 2) (3 2)

9 2

+ + −

=

− 6

Try This 2

Simplify (i) 3

12 (ii)

1

3+ 7 (iii)

7 2

7 2

+ −

Answers to Try This

1. (i) 3 3 2. (i) 3

2

(ii) 7 (ii) 7 3

4 −

(iii) 50 2 (iii) 11 4 7

Indices

If a positive integer a is multiplied by itself three times, we get a , i.e. 3 a a a× × =a3. Here a is

called the base and 3, the index or power. Thus a means the 44 th power of a.

In general, a means the nth power of a, where n is any positive index of the positive integer a. n

Rules of Indices

Example 1

Evaluate each of the following without using a calculator

(i) 7−1 (ii) 17 0 (iii)

Solving Exponential Equations Example 3

Solve the following exponential equations

(i) 2x =32 (ii) 4x 1+ =0.25

Solution:

(i) 2x =32 (ii) 4x 1+ =0.25

5

2x =2 4 1 1

4 x + ₌

5

x

∴ = 1 1

4x + =4−

1 1

x+ = −

∴ = −x 2

Try This 3

Solve the following equations:

(i) 3x =81 (ii) 32x =8 (iii) 7 1 49 x ₌

(iv) 5x =1 (v) 34x =27x 3+ (vi) 4x×32x =6 Example 4

Solve the equation 22 3x+ +2x 3+ = +1 2x.

Solution:

22 3x+ +2x 3+ = +1 2x

3 3

2x× × + × = +2x 2 2x 2 1 2x Let y=2x

8y2+8y= +1 y

8y2+7y− =1 0 (8y−1)(y+ =1) 0 1 8

y= or 1−

When 1

8

y= when y= −1

1 2

8

x _{= 2}

1

x _{= − (inadmissible)}

3

2x =2− 3

x

Try This 4

Solve the equation 32 1x+ + =9 3x 3+ +3x.

Example 5

If 3x×92y =27 and 2 4 1 8 x_× −y ₌

, calculate the values of x and y.

Solution: 2

3x×9 y =27 ⋯⋯⋯(1) 1

2 4 (2)

8 x_× −y ₌

⋯⋯⋯

From (1): 3x×(3 )2 2y =33 3x×34y =33 3x 4+ y =33

x+4y=3 ⋯⋯(3)

From (2): 2 (2 )2 1₃ 2 x_× −y ₌

2x×2−2y =2−3 2x 2− y =2−3

x−2y= −3 ⋯⋯(4) (3) (4) :− 6y=6

y=1

Substitute y=1 into (3): x+4(1)=3 x= −1 1

x

∴ = − and y=1.

Try This 5

Answers to Try This

1. (i) 1

7 (ii) 1 (iii) 343 (iv)

1 4

(v) 27 (vi) 3 (vii) 81 (viii) 16

2. (i) a 9 (ii) 4

a (iii)

5 6 a b

3. (i) x=4 (ii) 3

5

x= (iii) x= −2

(iv) x=0 (v) x=9 (vi) 1

2

x=

Logarithms

Definition:

For any number y such that y=ax (a>0 and a≠1), the logarithm of y to the base a is defined to be x and is denoted by log_a y.

Thus if y=ax, then log_a y=x

For example, 81 3= 4 ∴log 81₃ =4

2

100 10= ∴log 100₁₀ =2 Note: The logarithm of 1 to any base is 0, i.e. log 1_a =0.

The logarithm of a number to a base of the same number is 1, i.e. log_aa=1. The logarithm of a negative number is undefined.

Example 1

Find the value of (i) log 64 ₂ (ii) log 3 ₉ (iii) log₃1

9 (iv) log 0.25 8

Solution:

(i) Let log 64₂ =x (ii) Let log 3₉ =x

64=2x 3=9x

6

2 =2x 3=32x

6

x

∴ = 1=2x

1 2

x

∴ =

(iii) Let log₃1

9=x (iv) Let log 0.258 =x 1

3 9

x

= 0.25=8x

3−2 =3x 1 23 4

x =

∴ = −x 2 2−2 =23x

3x= −2

2

3

x

Laws of Logarithms

Without using tables, evaluate log₁₀ 41 log 70 log_{10 10}41 2 log 5₁₀

35+ − 2 + .

Changing the Base of Logarithms

Logarithms to base 10 such as log 5 and ₁₀ log (₁₀ x+1) are called common logarithms. An alternative form of writing log 5 is lg 5 .Common logarithms can be evaluated using a 10

scientific calculator.

Example 3

Find the value of log 16 . ₅

Solution:

10 5

10

log 16 1.204

log 16 1.722

log 5 0.699

= = =

Try This 2

Find the value of log 54 . ₄

Solving Logarithmic Equations

Example 4

Solve the equation 3x =18.

Solution: 3x =18

Taking logarithms to base 10 on both sides,

10 10

log 3x =log 18

10 10

log 3 log 18

x =

10

10

log 18 1.2553 log 3 0.4771

x= =

2.631 =

Try This 3

Solve the equation 5x 1+ =30. Example 5

Given that log 4 2 log₁₀ + ₁₀ p=2, calculate the value of p without using tables or calculators. Solution:

10 10

log 4 2 log+ p=2 log (4₁₀ ×p2)=2 4p2=102

2 100

4

p =

2

25

p =

5

p= ±

Try This 4

Solve the equation log₂ 3 1 3

2 7

x

x−+ = .

Example 6

Solve the equation log (3₁₀ x+ −2) 2 log₁₀x= −1 log (5₁₀ x−3).

Solution:

10 10 10

log (3x+ −2) 2 log x= −1 log (5x−3)

2

10 10 10

log (3x+ −2) log x +log (5x− =3) 1

10 2

(3 2)(5 3)

log x x 1

x

+ _{− =}

(3x 2)(5₂ x 3) 101

x

+ _{− =}

15x2−9x+10x− =6 10x2

5x2+ − =x 6 0 (5x+6)(x− =1) 0

6

5

x

∴ = − or x=1

Since x cannot be negative, x=1.

Try This 5

Solve the equation log₂x2= +4 log (₂ x−3).

Answers to Try This 1. log 160 ₃ 2. 2.877 3. x=1.113