EXPONENT RULES -
SIMPLIFYING OR PROVING BY
APPLYING LAWS OF EXPONENTS

Please study
RATIONAL EXPONENTS BEFORE EXPONENT RULES
if you have not already done so.

It is a prerequisite here.

There, we provided the explanation
for Rational Exponents.

We discussed how we can apply the same
7 Laws and the 2 Rules given for
whole number Exponents can be applied
for Fractional Exponents.

Here, we apply the Laws to simplify
a given expression or to prove a
given relationship, by applying
Laws of Exponents.

Solved Example 1 of Exponent Rules

Simplify (3ⁿ x 9^{n + 1})⁄(3^{n - 1} x 9^{n - 1})

Solution to Example 1 of Exponent Rules:

Let A = (3ⁿ x 9^{n + 1})⁄(3^{n - 1} x 9^{n - 1})
Writing powers with same base at one place, we get
A = {(3ⁿ)⁄(3^{n - 1})} x {(9^{n + 1})⁄(9^{n - 1})}
We know a^m⁄aⁿ = a^{m - n}
Applying this here, we get
A = {3^{n - (n - 1)}} x {9^{(n + 1) - (n - 1)}}
= {3^{n - n + 1)}} x {9^{(n + 1 - n + 1)}}
= (3¹) x (9²) = 3 x 9 x 9 = 243. Ans.

Solved Example 2 of Exponent Rules

Simplify { 3 x (27)ⁿ⁺¹ + 9 x (3)^3n-1}⁄{ 8 x 3³ⁿ - 5 x (27)ⁿ}

Solution to Example 2 :

Let A = { 3 x (27)ⁿ⁺¹ + 9 x (3)^3n-1}⁄{ 8 x 3³ⁿ - 5 x (27)ⁿ}
We know a^{m + n} = a^m x aⁿ
Applying this here, we get
A = { 3 x (27)ⁿ x (27)¹ + 9 x (3)³ⁿ x (3)^-1}⁄{ 8 x 3³ⁿ - 5 x (27)ⁿ}
We know 3³ⁿ = (3³)ⁿ = (27)ⁿ
Using this here, we get
A = { 3 x (27)ⁿ x (27) + 9 x (27)ⁿ x (1⁄3}⁄{ 8 x (27)ⁿ - 5 x (27)ⁿ}
We can see (27)ⁿ being present in each term. Taking it common, we get
A = [(27)ⁿ { 3 x 27 + 9 x (1⁄3}]⁄[(27)ⁿ{8 - 5 }]
Cancelling (27)ⁿ which is present in numerator and denominator, we get
A = ( 81 + 3)⁄(3) = 84⁄3 = 28. Ans.

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Solved Example 3 of Exponent Rules

Example 3 of Exponent Rules :

Simplify (x^1⁄3 - x^-1⁄3)(x^2⁄3 + 1 + x^-2⁄3)

Solution to Example 3 :

Let A = (x^1⁄3 - x^-1⁄3)(x^2⁄3 + 1 + x^-2⁄3)
We know x^2⁄3 = x^{2 x 1⁄3} = x^{(1⁄3) x 2} = {x^(1⁄3)}²
Similarly, x^-2⁄3 = {x^(-1⁄3)}²
If we denote x^1⁄3 by a, and x^-1⁄3 by b,
then ab = x^1⁄3 x x^-1⁄3 = 1 [Since pⁿ and p^-n are reciprocals to one another.]
Thus A becomes (a - b)(a² + ab + b²).
We know (a - b)(a² + ab + b²) = a³ - b³
∴ A = a³ - b³ = (x^1⁄3)³ - (x^-1⁄3)³
= x^{1⁄3 x 3} - x^{-1⁄3 x 3} = x¹ - x^-1 = x - 1⁄x. Ans.

Solved Example 4 of Exponent Rules

Example 4 of Exponent Rules :

If abc = 1,
prove that 1⁄(1 + a + b^-1) + 1⁄(1 + b + c^-1) + 1⁄(1 + c + a^-1) = 1.

Solution to Example 4 of Exponent Rules:

L.H.S. = 1⁄(1 + a + b^-1) + 1⁄(1 + b + c^-1) + 1⁄(1 + c + a^-1)
= 1⁄(1 + a + 1⁄b) + 1⁄(1 + b + 1⁄c) + 1⁄(1 + c + 1⁄a)
To prove, let us change the L.H.S. to two variables (from three),
by using the relation abc = 1. Let us eliminate c.
abc = 1 ⇒ c = 1⁄(ab); and c^-1 = 1⁄c = ab;
Substituting these in the L.H.S., we get
L.H.S. = 1⁄(1 + a + 1⁄b) + 1⁄(1 + b + ab) + 1⁄(1 + 1⁄(ab) + 1⁄a)
Multiplying the numerator and denominator of first term with b,
of third term with ab, we get
L.H.S. = (b x 1)⁄{b x (1 + a + 1⁄b)} + 1⁄(1 + b + ab)
+ (ab x 1)⁄{ab x (1 + 1⁄(ab) + 1⁄a)}
= (b)⁄{b x 1 + b x a + b x (1⁄b)} + 1⁄(1 + b + ab)
+ (ab)⁄[ab x 1 + ab x {1⁄(ab)} + ab x (1⁄a)]
= (b)⁄(b + ab + 1) + 1⁄(1 + b + ab) + (ab)⁄[ab + 1 + b ]
= (b + 1 + ab)⁄(b + ab + 1) = 1 = R.H.S. (proved.)

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Exercise on Exponent Rules

Solve the following problems on Exponent Rules.

1. Simplify: {5 x (25)ⁿ⁺¹ - 25 x (5)²ⁿ}⁄{ 5 x 5^{(2n + 3)} - (25)^{n + 1}}

2. Solve: √(a⁄b) = (b⁄a)^{1 - 3x}

3. Solve: {(2⁄3)^1⁄3}^{(x - 1)} = 27⁄8.

For Answers, see at the bottom of the page.

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Answers to Exercise on Exponent Rules

1. 1⁄6.
2. 1⁄2.
3. -8.