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 solve
problems with emphasis on the law
exponent of a power to some base
= multiply exponents to that base.

Solved Example 1 of Multiply Exponents

Simplify {(27)^{2n⁄3} x (8)^{-n⁄6}}⁄{(18)^{-n⁄2}}

Solution to Example 1 of Multiply Exponents:

Let A = {(27)^{2n⁄3} x (8)^{-n⁄6}}⁄{(18)^{-n⁄2}}
We know 27 = 3 x 3 x3 = 3^{3}; 8 = 2 x 2 x 2 = 2^{3}; 18 = 2 x 9 = 2 x 3 x 3 = 2 x 3^{2}
∴ A = {(3^{3})^{2n⁄3} x (2^{3})^{-n⁄6}}⁄{(2 x 3^{2})^{-n⁄2}}
We Know
(a^{m})^{n} = a^{mn}]
(ab)^{m} = a^{m} x b^{m}
Applying these, we get
A = {(3^{3 x 2n⁄3}) x (2^{3 x -n⁄6})}⁄{(2^{-n⁄2}) x (3^{2})^{-n⁄2}}
= {(3^{2n}) x (2^{-n⁄2})}⁄{(2^{-n⁄2}) x (3^{2 x -n⁄2})}
= (3^{2n})⁄(3^{-n}) [By cancelling (2^{-n⁄2}) which is in Numerator and denominator]
= (3^{2n - (-n)}) = (3^{2n + n}) = 3^{3n}. Ans.

We know a^{m}⁄a^{n} = a^{m - n}
And this can be applied to fractional Exponents also.
Applying this here, we get
L.H.S. = (x^{a - b})^{(a + b)}.(x^{b - c})^{(b + c)}.(x^{c - a})^{(c + a)}.
We Know
(a^{m})^{n} = a^{mn}
And this can be applied to fractional Exponents also.
Applying this here, we get
L.H.S. = {x^{(a - b)(a + b)}}.{x^{(b - c)(b + c)}}.{x^{(c - a)(c + a)}}
We know (a - b)(a + b) = a^{2} - b^{2}; (b - c)(b + c) = b^{2} - c^{2}; (c - a)(c + a) = c^{2} - a^{2};
∴ L.H.S. = {x^{(a2 - b2)}}.{x^{(b2 - c2)}}.{x^{(c2 - a2)}}
We Know a^{m} x a^{n} = a^{m + n}
And this can be applied to fractional Exponents also.
Applying this here, we get
L.H.S. = x^{{(a2 - b2) + (b2 - c2) + (c2 - a2)}}.
= x^{{(a2 - a2) + (b2 - b2) + (c2 - c2)}}.
= x^{{0 + 0 + 0}}. = x^{0} = 1 = R.H.S. (proved)

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If a^{x} = b, b^{y} = c, c^{z} = a, show that xyz = 1.

Solution to Example 3 of Multiply Exponents:

a^{x} = b .........(i); b^{y} = c .........(ii); c^{z} = a.........(iii)
Using (iii) in (i), we get a^{x} = b ⇒ (c^{z})^{x} = b ⇒ c^{zx} = b [Since (a^{m})^{n} = a^{mn}]
Using (ii) here, we get c^{zx} = b ⇒ (b^{y})^{zx} = b ⇒ b^{yzx} = b
⇒ b^{xyz} = b^{1} ⇒ xyz = 1. (Proved.)

By data (64)^{x} = 1⁄(256)^{y}
We can see that 64 and 256 are powers of 4.
64 = 4 x 4 x 4 = 4^{3}; 256 = 4 x 4 x 4 x 4 = 4^{4};
(64)^{x} = 1⁄(256)^{y} ⇒ (64)^{x} = (256)^{-y} ⇒ (4^{3})^{x} = (4^{4})^{-y}
We Know
(a^{m})^{n} = a^{mn}
Applying this here, we get
4^{3x} = 4^{4 x -y} ⇒ 4^{3x} = 4^{-4y}
Since the bases are same, the exponents have to be equal.
∴ 3x = -4y ⇒ 3x + 4y = 0. (Proved.)

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