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

Please study
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.

Here, in Addition of Exponents,
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 a^m x aⁿ = a^{m + n}
Applying this Law, we get
A = (256 x 2) x (x^{2 + 4}) x {y^{3 + (-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 = a^m x b^m (See Law 6)
Applying this Law, we get
A = (2x²y) x (-3)³(x²)³(y²)³ x (x)^-1(y)^-1
We know (a^m)ⁿ = a^mn (See Law 2)
Applying this Law, we get
A = (2x²y) x (-3) x (-3) x (-3) (x^{2 x 3})(y^{2 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)(x^{2 x 3})(x^-1)(y )(y^{2 x 3})(y^-1)
⇒ A = (2) x (-27)(x²)(x⁶)(x^-1)(y )(y⁶)(y^-1)
We know a^m x aⁿ = a^{m + n}
Applying this Law, we get
A = (-54){x^{2 + 6 + (-1)}}{y^{1 + 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)(a⁹⁄a⁷)(b⁴⁄b²)(c²⁄c)
We know a^m⁄aⁿ = a^{m - n}
Applying this here, we get
A = (3)(a^{9 - 7})(b^{4 - 2})(c^{2 - 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 = a^m x b^m (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 (a^m)ⁿ = a^mn (See Law 2)
Applying this Law, we get
A = {9(a^{2 x 2})(b^{3 x 2})(c^{4 x 2})}⁄{4(a^{4 x 2})(b^{3 x 2})(c^{2 x 2})}
⇒ A = (9a⁴b⁶c⁸)⁄(4a⁸b⁶c⁴)
Grouping constant terms and variables of one type at one place, we get
A = (9⁄4)(a⁴⁄a⁸)(b⁶⁄b⁶)(c⁸⁄c⁴)
We know a^m⁄aⁿ = a^{m - n}
Applying this here, we get
A = (9⁄4)(a^{4 - 8})(b^{6 - 6})(c^{8 - 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 a^{x(y - z)} x a^{y(z - x)} x a^{z(x - y)} = 1.

Solution to Example 5 of Addition of Exponents :

We know a^m x aⁿ = a^{m + n}
and we can apply this Law for 3 terms also.
∴ L.H.S. = a^{x(y - z)} x a^{y(z - x)} x a^{z(x - y)}
= a^{x(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 (x^a⁄x^b)^c x (x^b⁄x^c)^a x (x^c⁄x^a)^b

Solution to Example 6 of Addition of Exponents :

Let A = (x^a⁄x^b)^c x (x^b⁄x^c)^a x (x^c⁄x^a)^b
We know a^m⁄aⁿ = a^{m - n}
Applying this Law, we get
A = (x^{a - b})^c x (x^{b - c})^a x (x^{c - a})^b
We know (a^m)ⁿ = a^mn
Applying this Law, we get
A = x^{(a - b)c} x x^{(b - c)a} x x^{(c - a)b}
We know a^m x aⁿ = a^{m + 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 (x^a⁄x^b)^c x (x^b⁄x^c)^a x (x^c⁄x^a)^b = 1. Ans.

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

Exercise : 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.

Answers to Exercise : Addition of Exponents

1. (i) 25 (ii) 2 (iii) 1024 (iv) 3⁸⁄2²
2. (i) 27 (ii) 1⁄16 (iii) 125 (iv) 81
   
3. (i) 3⁴⁸ (ii) 7¹⁵⁄(-5)¹² (iii) 1⁄256 (iv) 1⁄16
4. (i) 15 (ii) 4⁄25 (iii) 10000 (iv) (m⁵ x z⁶)⁄(n⁵ x l³)