RULES OF EXPONENTS -
PROVING A RELATIONSHIP BY
APPLYING LAWS OF EXPONENTS

Please study
RATIONAL EXPONENTS BEFORE RULES OF EXPONENTS
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, 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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Solved Example 1 of Rules of Exponents

If a + b + c = 0, show that x^(a2b-1c-1).x^(a-1b2c-1).x^(a-1b-1c2) = x³.

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 (a²b^-1c^-1) with abc, we get
(a²b^-1c^-1) = abc(a²b^-1c^-1)⁄abc = (a.a² b.b^-1c.c^-1)⁄abc = (a³.1.1)⁄abc [Since b.b^-1 = 1 = c.c^-1]
= (a³)⁄(abc).
Similarly, (a^-1b²c^-1) = (b³)⁄(abc).
and (a^-1b^-1c²) = (c³)⁄(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 aⁿ = 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, a³ + b³ + c³ = 3abc

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Solved Example 2 of Rules of Exponents

If a = x + √(x² + 1), show that x = (a - a^-1)⁄2.

Solution to Example 2 of Rules of Exponents:

a = x + √(x² + 1) ⇒a^-1 = 1⁄{x + √(x² + 1)}
Multiplying numerator and denominator with {x - √(x² + 1)}, we get
a^-1 = 1{x - √(x² + 1)}⁄[{x + √(x² + 1)}{x - √(x² + 1)}]
We know (a + b)(a - b) = (a² - b²)
Applying this here, we get
a^-1 = {x - √(x² + 1)}⁄[x² - {√(x² + 1)}²]
= {x - √(x² + 1)}⁄[x² - {x² + 1}]
= {x - √(x² + 1)}⁄[x² - x² - 1]
= {x - √(x² + 1)}⁄[-1]
= - {x - √(x² + 1)}
a - a^-1 = {x + √(x² + 1)} - [- {x - √(x² + 1)}]
= {x + √(x² + 1)} + {x - √(x² + 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⁄aⁿ
Applying this here, we get
a^{x - 1} = bc ⇒ a^x⁄a¹ = bc ⇒ a^x⁄a = bc ⇒ a^x = abc............(i)
b^{y - 1} = ca ⇒ b^y⁄b¹ = ca ⇒ b^y⁄b = ca ⇒ b^y = bca = abc............(ii)
c^{z - 1} = ab ⇒ c^z⁄c¹ = 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 aⁿ = 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.

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Exercise on Rules of Exponents

Sove the following problems
on Rules of Exponents.

2. If 2^x = 3^y = 6^-z, show that: 1⁄x + 1⁄y + 1⁄z = 0.
3. If (9ⁿ x 3² x 3ⁿ - 27ⁿ)⁄(3^3m x 2³) = 3^-3, Prove that (m - n) = 1.

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