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

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
am x an = am + 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

Proof:
a + b + c = 0 ⇒ a + b = -c ⇒ (a + b)3 = (-c)3
a3 + b3 + 3ab(a + b) = -c3a3 + b3 + 3ab(-c) = -c3
a3 + b3 - 3abc = -c3a3 + b3 + c3 = 3abc (proved.)
Applying this here, we get
L.H.S. = x(3abc)⁄(abc) = x3 = R.H.S. (proved.)

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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 ax - 1 = bc; by - 1 = ca; cz - 1 = ab, show that xy + yz + zx = xyz

Solution to Example 3 of Rules of Exponents :

We know am - n = aman
Applying this here, we get
ax - 1 = bcaxa1 = bcaxa = bcax = abc............(i)
by - 1 = cabyb1 = cabyb = caby = bca = abc............(ii)
cz - 1 = abczc1 = abczc = abcz = cab = abc...........(iii)
Raising power yz to (i), zx to (ii), xy to (iii), we get
(ax)yz = (abc)yz ⇒ (axyz) = (abc)yz...........(iv)
(by)zx = (abc)zx ⇒ (bxyz) = (abc)zx...........(v)
(cz)xy = (abc)xy ⇒ (cxyz) = (abc)xy...........(vi)
Multiplying (iv), (v) and (vi), we get
(axyz) x (bxyz) x (cxyz) = (abc)yz x (abc)zx x (abc)xy
Applying the Law (am)(bm) = (ab)m to the L.H.S. and
the Law am x an = am + 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.

1. If 2x = 3y = 6-z, show that: 1⁄x + 1⁄y + 1⁄z = 0.
2. If (9n x 32 x 3n - 27n)⁄(33m x 23) = 3-3, Prove that (m - n) = 1.

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