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, in Rules of Exponents,
we apply the Laws to prove
a given relationship,using the
given condition, by applying
Laws of Exponents.

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If a + b + c = 0, show that x^{(a2b-1c-1)}.x^{(a-1b2c-1)}.x^{(a-1b-1c2)} = x^{3}.

Solution to Example 1 of Rules of Exponents:

L.H.S. = x^{(a2b-1c-1)}.x^{(a-1b2c-1)}.x^{(a-1b-1c2)}
Multipying and dividing (a^{2}b^{-1}c^{-1}) with abc, we get
(a^{2}b^{-1}c^{-1}) = abc(a^{2}b^{-1}c^{-1})⁄abc = (a.a^{2}b.b^{-1}c.c^{-1})⁄abc = (a^{3}.1.1)⁄abc [Since b.b^{-1} = 1 = c.c^{-1}]
= (a^{3})⁄(abc).
Similarly, (a^{-1}b^{2}c^{-1}) = (b^{3})⁄(abc).
and (a^{-1}b^{-1}c^{2}) = (c^{3})⁄(abc)
Substituting these in L.H.S., we get
L.H.S. = x^{(a3)⁄(abc)}.x^{(b3)⁄(abc)}.x^{(c3)⁄(abc)}
We Know a^{m} x a^{n} = a^{m + n}
Applying this here, we get
L.H.S. = x^{(a3)⁄(abc) + (b3)⁄(abc) + (c3)⁄(abc)} = x^{(a3 + b3 + c3)⁄(abc)}
We have

If a + b + c = 0, a^{3} + b^{3} + c^{3} = 3abc

Proof: a + b + c = 0 ⇒ a + b = -c ⇒ (a + b)^{3} = (-c)^{3}
⇒ a^{3} + b^{3} + 3ab(a + b) = -c^{3} ⇒ a^{3} + b^{3} + 3ab(-c) = -c^{3}
⇒ a^{3} + b^{3} - 3abc = -c^{3} ⇒ a^{3} + b^{3} + c^{3} = 3abc (proved.)
Applying this here, we get
L.H.S. = x^{(3abc)⁄(abc)} = x^{3} = R.H.S. (proved.)

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If a = x + √(x^{2} + 1), show that x = (a - a^{-1})⁄2.

Solution to Example 2 of Rules of Exponents:

a = x + √(x^{2} + 1) ⇒a^{-1} = 1⁄{x + √(x^{2} + 1)}
Multiplying numerator and denominator with {x - √(x^{2} + 1)}, we get a^{-1} = 1{x - √(x^{2} + 1)}⁄[{x + √(x^{2} + 1)}{x - √(x^{2} + 1)}]
We know (a + b)(a - b) = (a^{2} - b^{2})
Applying this here, we get a^{-1} = {x - √(x^{2} + 1)}⁄[x^{2} - {√(x^{2} + 1)}^{2}]
= {x - √(x^{2} + 1)}⁄[x^{2} - {x^{2} + 1}]
= {x - √(x^{2} + 1)}⁄[x^{2} - x^{2} - 1]
= {x - √(x^{2} + 1)}⁄[-1]
= - {x - √(x^{2} + 1)} a - a^{-1} = {x + √(x^{2} + 1)} - [- {x - √(x^{2} + 1)}]
= {x + √(x^{2} + 1)} + {x - √(x^{2} + 1)}
= {x} + {x} = 2x ⇒ (a - a^{-1})⁄2 = x
Thus x = (a - a^{-1})⁄2 (proved.)

Solved Example 3 of Rules of Exponents

If a^{x - 1} = bc; b^{y - 1} = ca; c^{z - 1} = ab, show that xy + yz + zx = xyz

Solution to Example 3 of Rules of Exponents :

We know a^{m - n} = a^{m}⁄a^{n} Applying this here, we get a^{x - 1} = bc ⇒ a^{x}⁄a^{1} = bc ⇒ a^{x}⁄a = bc ⇒ a^{x} = abc............(i) b^{y - 1} = ca ⇒ b^{y}⁄b^{1} = ca ⇒ b^{y}⁄b = ca ⇒ b^{y} = bca = abc............(ii) c^{z - 1} = ab ⇒ c^{z}⁄c^{1} = ab ⇒ c^{z}⁄c = ab ⇒ c^{z} = cab = abc...........(iii) Raising power yz to (i), zx to (ii), xy to (iii), we get (a^{x})^{yz} = (abc)^{yz} ⇒ (a^{xyz}) = (abc)^{yz}...........(iv) (b^{y})^{zx} = (abc)^{zx} ⇒ (b^{xyz}) = (abc)^{zx}...........(v) (c^{z})^{xy} = (abc)^{xy} ⇒ (c^{xyz}) = (abc)^{xy}...........(vi) Multiplying (iv), (v) and (vi), we get (a^{xyz}) x (b^{xyz}) x (c^{xyz}) = (abc)^{yz} x (abc)^{zx} x (abc)^{xy} Applying the Law (a^{m})(b^{m}) = (ab)^{m} to the L.H.S. and the Law a^{m} x a^{n} = a^{m + n} to the R.H.S., we get (abc)^{xyz} = (abc)^{yz + zx + xy} Applying one of the Rules of Exponents: Since the bases are same, the exponents have to be equal. ∴ xyz = yz + zx + xy ⇒ xy + yz + zx = xyz (Proved.) Thus the problem on Rules of Exponents is solved.

Sove the following problems
on Rules of Exponents.

If 2^{x} = 3^{y} = 6^{-z}, show that: 1⁄x + 1⁄y + 1⁄z = 0.

If (9^{n} x 3^{2} x 3^{n} - 27^{n})⁄(3^{3m} x 2^{3}) = 3^{-3}, Prove that (m - n) = 1.

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