# EXPONENT RULES - SIMPLIFYING OR PROVING BY APPLYING LAWS OF EXPONENTS

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 9n + 1)⁄(3n - 1 x 9n - 1)

Solution to Example 1 of Exponent Rules:

Let A = (3n x 9n + 1)⁄(3n - 1 x 9n - 1)
Writing powers with same base at one place, we get
A = {(3n)⁄(3n - 1)} x {(9n + 1)⁄(9n - 1)}
We know aman = am - n
Applying this here, we get
A = {3n - (n - 1)} x {9(n + 1) - (n - 1)}
= {3n - 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 am + n = am 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 (x1⁄3 - x-1⁄3)(x2⁄3 + 1 + x-2⁄3)

Solution to Example 3 :

Let A = (x1⁄3 - x-1⁄3)(x2⁄3 + 1 + x-2⁄3)
We know x2⁄3 = x2 x 1⁄3 = x(1⁄3) x 2 = {x(1⁄3)}2
Similarly, x-2⁄3 = {x(-1⁄3)}2
If we denote x1⁄3 by a, and x-1⁄3 by b,
then ab = x1⁄3 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 = (x1⁄3)3 - (x-1⁄3)3
= x1⁄3 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: √(ab) = (ba)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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