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Please study
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 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 = 33; 8 = 2 x 2 x 2 = 23; 18 = 2 x 9 = 2 x 3 x 3 = 2 x 32
∴ A = {(33)2n⁄3 x (23)-n⁄6}⁄{(2 x 32)-n⁄2}
We Know
(am)n = amn]
(ab)m = am x bm
Applying these, we get
A = {(33 x 2n⁄3) x (23 x -n⁄6)}⁄{(2-n⁄2) x (32)-n⁄2}
= {(32n) x (2-n⁄2)}⁄{(2-n⁄2) x (32 x -n⁄2)}
= (32n)⁄(3-n) [By cancelling (2-n⁄2) which is in Numerator and denominator]
= (32n - (-n)) = (32n + n) = 33n. Ans.

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Solved Example 2 of Multiply Exponents

Prove that (xaxb)(a + b).(xbxc)(b + c).(xcxa)(c + a) = 1.

Solution to Example 2 of Multiply Exponents:

We know aman = am - n
And this can be applied to fractional Exponents also.
Applying this here, we get
L.H.S. = (xa - b)(a + b).(xb - c)(b + c).(xc - a)(c + a).
We Know
(am)n = amn
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) = a2 - b2; (b - c)(b + c) = b2 - c2; (c - a)(c + a) = c2 - a2;
∴ L.H.S. = {x(a2 - b2)}.{x(b2 - c2)}.{x(c2 - a2)}
We Know
am x an = am + 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}. = x0 = 1 = R.H.S. (proved)

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Solved Example 3 of Multiply Exponents

If ax = b, by = c, cz = a, show that xyz = 1.

Solution to Example 3 of Multiply Exponents:

ax = b .........(i); by = c .........(ii); cz = a.........(iii)
Using (iii) in (i), we get
ax = b ⇒ (cz)x = bczx = b [Since (am)n = amn]
Using (ii) here, we get
czx = b ⇒ (by)zx = bbyzx = b
bxyz = b1xyz = 1. (Proved.)

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Solved Example 4 of Multiply Exponents

If (64)x = 1⁄(256)y, show that 3x + 4y = 0.

Solution to Example 4 of Multiply Exponents:

By data (64)x = 1⁄(256)y
We can see that 64 and 256 are powers of 4.
64 = 4 x 4 x 4 = 43; 256 = 4 x 4 x 4 x 4 = 44;
(64)x = 1⁄(256)y ⇒ (64)x = (256)-y ⇒ (43)x = (44)-y
We Know
(am)n = amn
Applying this here, we get
43x = 44 x -y ⇒ 43x = 4-4y
Since the bases are same, the exponents have to be equal.
∴ 3x = -4y ⇒ 3x + 4y = 0. (Proved.)

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Exercise on Multiply Exponents

  1. Simplify: {7(2n + 3) - (49)(n + 2)}⁄{(343)(n + 1)}2⁄3
  2. (243)-3⁄5

For Answers, see at the bottom of the page.

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Answers to Exercise on Multiply Exponents

  1. -42.