# ADDITION OF EXPONENTS - PRODUCT/QUOTIENT OF POWERS = ADDITION/DIFFERENCE OF EXPONENTS APPLIED IN EXAMPLES, EXERCISE

Laws of Exponents before Addition of Exponents,
if you have not already done so.

It is a prerequisite here.

There, we stated the 7 laws of indices
and the two Rules used for solving problems.

We also gave Links to the explanations
and proofs of the 7 laws.

we apply the two Laws of Exponents
where product/quotient of powers
of the same base = addition/difference
of exponents to that base.
Solved Examples, Exercise, provided,
help in mastering the Laws.

### Solved Example 1 : Addition of Exponents

Simplify:
x2 x (-4x)4 x y3 x 2y-2

Solution to Example 1 of Addition of Exponents :

Let A = x2 x (-4x)4 x y3 x 2y-2
We know (-4x)4 = (-4)4 x x4.
Grouping constant terms and variables of one type at one place, we get
A = {(-4)4 x 2} x (x2 x x4) x (y3 x y-2)
We know even power of minus will become plus.
So (-4)4 = 44 = 4 x 4 x 4 x 4 = 256.
We know am x an = am + n
Applying this Law, we get
A = (256 x 2) x (x2 + 4) x {y3 + (-2)} = 512 x x6 x y1 = 512x6y
Thus x2 x (-4x)4 x y3 x 2y-2 = 512x6y. Ans.

### Solved Example 2 : Addition of Exponents

Simplify:
(2x2y) x (-3x2y2)3 x (xy)-1

Solution to Example 2 of Addition of Exponents :

Let A = (2x2y) x (-3x2y2)3 x (xy)-1
We know (ab)m = am x bm (See Law 6)
Applying this Law, we get
A = (2x2y) x (-3)3(x2)3(y2)3 x (x)-1(y)-1
We know (am)n = amn (See Law 2)
Applying this Law, we get
A = (2x2y) x (-3) x (-3) x (-3) (x2 x 3)(y2 x 3) x (x)-1(y)-1
Grouping constant terms and variables of one type at one place, we get
A = (2) x (-3) x (-3) x (-3) (x2)(x2 x 3)(x-1)(y )(y2 x 3)(y-1)
⇒ A = (2) x (-27)(x2)(x6)(x-1)(y )(y6)(y-1)
We know am x an = am + n
Applying this Law, we get
A = (-54){x2 + 6 + (-1)}{y1 + 6 + (-1)}
⇒ A = (-54){x7}{y6}
Thus 2x2y) x (-3x2y2)3 x (xy)-1 = -54x7y6. Ans.

### Solved Example 3 : Addition of Exponents

Simplify:
(6a9b4c2)⁄(2a7b2c)

Solution to Example 3 of Addition of Exponents :

Let A = (6a9b4c2)⁄(2a7b2c)
Grouping constant terms and variables of one type at one place, we get
A = (6⁄2)(a9a7)(b4b2)(c2c)
We know aman = am - n
Applying this here, we get
A = (3)(a9 - 7)(b4 - 2)(c2 - 1) = 3a2b2c1
Thus (6a9b4c2)⁄(2a7b2c) = 3a2b2c. Ans.

### Solved Example 4 : Addition of Exponents

Simplify:
(3a2b3c4)2⁄(2a4b3c2)2

Solution to Example 4 of Addition of Exponents :

Let A = (3a2b3c4)2⁄(2a4b3c2)2
We know (ab)m = am x bm (See Law 6)
Applying this Law, we get
A = {32(a2)2(b3)2(c4)2}⁄{22(a4)2(b3)2(c2)2}
We know (am)n = amn (See Law 2)
Applying this Law, we get
A = {9(a2 x 2)(b3 x 2)(c4 x 2)}⁄{4(a4 x 2)(b3 x 2)(c2 x 2)}
⇒ A = (9a4b6c8)⁄(4a8b6c4)
Grouping constant terms and variables of one type at one place, we get
A = (9⁄4)(a4a8)(b6b6)(c8c4)
We know aman = am - n
Applying this here, we get
A = (9⁄4)(a4 - 8)(b6 - 6)(c8 - 4)
⇒ A = (9⁄4)(a-4)(b0)(c4)
We know a-4 = 1⁄a4 and b0 = 1.
∴ A = (9⁄4)(1⁄a4)(1)(c4)
⇒ A = (9c4)⁄(4a4)
Thus (3a2b3c4)2⁄(2a4b3c2)2 = (9c4)⁄(4a4). Ans.

### Solved Example 5 : Addition of Exponents

Simplify:
Show that ax(y - z) x ay(z - x) x az(x - y) = 1.

Solution to Example 5 of Addition of Exponents :

We know am x an = am + n
and we can apply this Law for 3 terms also.
∴ L.H.S. = ax(y - z) x ay(z - x) x az(x - y)
= ax(y - z) + y(z - x) + z(x - y)
= a(xy - xz + yz - yx + zx - zy)
Since xy = yx, yz = zy, zx = xz, we have
L.H.S = a(xy - xy + yz - yz + zx - zx)
= a(0 + 0 + 0) = a(0) = a0.
We know a0 = 1.
By applying this Law, we get
L.H.S. = 1 = R.H.S. (proved)

### Solved Example 6 : Addition of Exponents

Simplify:
Simplify (xaxb)c x (xbxc)a x (xcxa)b

Solution to Example 6 of Addition of Exponents :

Let A = (xaxb)c x (xbxc)a x (xcxa)b
We know aman = am - n
Applying this Law, we get
A = (xa - b)c x (xb - c)a x (xc - a)b
We know (am)n = amn
Applying this Law, we get
A = x(a - b)c x x(b - c)a x x(c - a)b
We know am x an = am + n
Applying this Law, we get
A = x(a - b)c + (b - c)a + (c - a)b
⇒ A = x(ac - bc) + (ba - ca) + (cb - ab)
regrouping, we get
A = x(ac - ca) + (ba - ab ) + (cb - bc)
Since ac = ca, ba = ab, cb = bc, we get
A = x(0) + (0) + (0) = x0
We know a0 = 1.
Applying this, we get
A = 1.
Thus (xaxb)c x (xbxc)a x (xcxa)b = 1. Ans.

For More Solved Examples on application
of Laws of Exponents, go to
Math Exponents.

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### Exercise : Addition of Exponents

Problems on Addition of Exponents :

1. Simplify: (i) 54 x 53 x 5-5 (ii) 30 x 43 x 2-5
(iii) (-2)4 x (-4)2 x 2-2 x 42 (iv) 36 x 62 x 2-4
2. Simplify: (i) 38⁄35 (ii) (25 x 24)⁄213
(iii) 1⁄5-3 (iv) (-3)-4⁄(-3)-8
3. Simplify: (i) {(32)4}6 (ii) {(-5)4 x 75}3
(iii) {(-8)-4 x 2-8}2 (iv) {(4)-2 x 6-2}2 x 3-4
4. Evaluate : (i) (3-3 x 5-2 x 36)⁄(25 x 3-2 x 5-5)
(ii) {(23)2 x (-5)2}⁄{(52)2 x 16}
(iii) {(2-3)2 x (32)-3 x (5-4)2 }⁄{(2-2)5 x (33)-2 x (5-3)4 }
(iv) (l-4 x m3 x n-2 x z6)⁄(l-1 x m-2 x n3 x z0)

For the Answers, see at the bottom of the page.

For more problems for practice on Application
of Laws of exponents, go to
Math Exponents.

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