Maths

Unit: 10 Indices

10.0 Review

 We have to multiply a number by itself several times in mathematics.
 If a number is multiplied by itself 4 times, we will write it as × × ×

= 4.
 Similarly, when x is multiplied by itself m times, we will write it x x x x x ....

m times = m
 In xm, is a real number and m is a positive integer. In m, occurs m times

as a repeated factor. So, m is called the index of and is called the base.
The index of a base is also called the exponent or power. The plural form of
index is indices.

Laws of Indices:

There are some laws of indices which are as follows:

i) Product law
am× an = am+n where a ≠ 0.

ii) Division law
am÷ an = am-n, where a≠ 0.

iii) Power law
(am)n = amn, where a ≠ 0.

= , where a ≠ 0, b ≠ 0

(a x b)m = am× bm, where a ≠ 0, b ≠ 0

iv) a –m = , where a ≠ 0

v) √ = , where a ≠ 0, m ∈ N.
vi) √ = ⁄ , where a ≠ 0, m, n ∈ N
vii) a0 =1, where a ≠ 0.

10.1 Simplification of Indices:

We have to simplify the problems of indices by using the above laws of indices.
Discuss the laws of indices in the group and solve the following problems in the
index form.

3 × 3 × 3 × 3 = ……..

2a × 2a × 2a× 2a × 2a × 2a =……
2× 3× 5× -6× 4 = . . . .
9÷ 4 = . . . .

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(-3)2× (-3)4× (-3)7× (- 3) -5= . . . .
( + 3) × ( + 3)3× ( + 3)-7× ( + 3)-4 = . . .

3× 2× -5 = . . . .

Example 1: (b) (27) × (8) ÷ (18)
Find the value of

(a)
Solution:

(a)

= = = ==

(b) (27) × (8) ÷ (18)

= (3 ) × (2 ) ×

()

= ( ) ×( )

( ×)

= ×( )

( ) ×( )

= ×( )
= 3 × 20 = 3 × 1 = 3

Example 2: Simplify:

(a) × (b) × ×

Solution: Here, = =2× =
(a) ×

=×

××

= ( )=

()

(b) × ×

=( ) ×( ) ×( )

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= ( )( )× ( )( ) ( )( )
×
=× × +
=
= a0
=1
Example 3: Simplify:

(a) (b) +

Solution:

(a)

=

=

=
=
=
==

(b) ++
= ++
=
.. + . .. + . ..
. .
. .
=
= + +

=1

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Example 4: =1
If x + y + z = 0, prove that:

++

Solution: + +
L.H.S = )+ (
+
= +( )
=+ +
[ + + = 0]
=+

=
= 1 R.H.S proved

Exercise 10.1 (c) ×
1. Evaluate the following:

(a) (b) (4) × 3 × 3 ÷ 9

(d) 4 − 2 × 7 − (e) × × × (c)
2. Simplify
×× ×
(a) ( 2 – y-2)÷ ( -1 – y-1)
(b) ×
(d) ×
()
×
(e) ( ) ÷
3. Simplify

(a) × ×

(b) × ×

(c) × × (d) ×

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4. Simplify: (b)
(a)

(c) (d) - ) +( )

( )(

5. Simplify: + +
(a)

(b) + +

(c) + +

6. If p = , q = and r = , prove that: pb-c× qc-a× ra-b =1

7. If abc = 1, prove that: + + =1

8. If a = q+ryp, b = p+r.yq and c= p+q.yr, prove that:
q-r. br-p. cp-q = 1

9. If 2xyz = 1 and 3 + y3 + z3 = 1. Prove that:
ax2 y-1 z-1× ay2.z-1.x-1×az2. x-1.y-1= a2

10. Make 2/2 questions similar to Q.No. 1, 2 and 3 and find the solution of those
questions discussing in the group.

10.2. Exponential Equation:

Let's see the following equations:
2x = 8 and 2 = 8.
What are the value of in both equations?
Are both the equations satisfied by the same value of ?
In equation 2x =8, is in the form of base. But in equation 2 = 8, is the index of
2. The equation 2 = 8 is an exponential equation. Some more examples of the
exponential equations are 2 – 2 = 4, 7 – 7 = 0, 2 = 4 , etc.
An exponential equation can be solved by using the following axioms.
If the bases of left and right hand sides of an equation are the same, their index
must be equal.
i.e. if = , then =y.

if = 1, then = 0.

130 Mathematics, grade 10

Example 1: Solve: (b)3 + 3 – 28 = 0

(a) 3 = 1 (b) 3 + 3 – 28 = 0
Or, 3 .33 +3 =28
Solution:
Or, 3 (27 + 1) = 28
(a) 3 = 1
Or, 3 = 30 Or, 3 × 28 = 28
∴ −1=0
i.e. = 1 Or, 3 = 1
Or, 3 = 30
∴ =0

Example 2 : Solve: 4 – 6× 2 + 2 =0

Solution: Here,

4 – 6× 2 + 2 =0
Or, (2 ) – 6 × 2 × 2-1 + 2= 0

Or, (2 )2 – 6 × 2 × + 2 = 0

Or, (2 )2 – 3 × 2 + 2 = 0

Let 2 = a, then equation is
a2 – 3a + 2 = 0

Or, a2 – 2a – a + 2 = 0

Or, a(a - 2) – 1(a - 2) = 0

Or, (a - 2) (a - 1) = 0

Either, Or,

a -2 = 0 a -1 = 0

or, a = 2 or, a = 1

or, 2 = 21 or, 2 = 2°
∴ =1 ∴ =0
∴ = 0, 1

Example 3

Solve: + = 16
Solution: Here,
4 + 4 = 16

4+ =

Let, 4 = y, then equation is y + =

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Or, 16y2 + 16 = 257y Or, 16y – 1 = 0
Or, 16y2 – 257y + 16 =0 Or, y =
Or, 16y2 – 256y – y + 16 = 0
Or, 16y (y - 16) -1 (y - 16) = 0 Or, 4 =
Or, (y -16) (16y -1) = 0 Or, 4 = 4
Either, y- 16 =0 ∴ = -2
Or, 4 = 16
Or, 4 = 4
∴ =2

   x=±2

Exercise 10.2

1. Solve: (b) 5 = 1 (c) ÷ =1
(a) 3 = 27 (e) 2 = √2
(d) 2 = 16

2. Solve:

(a)2 + 2 = 3 (b) 2 + 2 = 9 (c) 3 + 3 =

(d) 2 + 2 = 66 (e) 2 = 4

3. Solve: (b) 9x - 12 x 3x -1 + 3 = 0
(a) 4x - 10 x 2x -1 + 4 = 0 (d) 16y - 5 x 4y+1 + 64 = 0
(c) 9x - 6 x 3x -1 = 3 (f) 22y+3 - 9 x 2y + 1 = 0
(e) 32x - 4 x 3x + 1 + 27 = 0

4. Solve: (b) 4x + = 4 (c) 7x + 343 x 7-x = 56
(a) 2x + 2-x = 2

(d) 3x + 3-x = 9 (e) 5x + 2 = 25 (f) 5x+1 + 52–x = 126

5. Make three equations of indices and give to your friends to find the solution. After
that present the solution to the class.

132 Mathematics, grade 10