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Question

Mathematics Question on Laws of Exponents

Evaluate

  1. (8−1×53)2−4\frac{ (8^{−1}× 5^3)}{2^{−4}}
  2. (5−1×2−1)×6−1(5^{−1}× 2^{−1})×6^{−1}
Answer

(i) (8−1×53)2−4\frac{(8^{−1} × 5^3) }{ 2^{−4}}
Since, a−m=1am,am÷an=am−na^{−m} = \frac{1}{a^m},a^m÷a^n = a^{m−n}
(8−1×53)2−4\frac{(8^{−1} × 5^3) }{ 2^{−4}}

=(24×53)81= \frac{(2^4 × 5^3)}{8^1} [Since a−m=1ama^{−m} = \frac{1}{a^m}]

=(24×53)23= \frac{(2^4 × 5^3)}{2^3}
=24−3×53= 2^{4 − 3} × 5^3 [am ÷ an = am − n]
= 2 × 125
= 250

(ii) (5−1×2−1)×6−1(5^{−1}× 2^{−1}) × 6^{−1}
Since,am×bm=(ab)m,a−m=1am a^m × b^m = (ab)^m, a^{-m} = \frac{1}{a^m}
(5−1×2−1)×6−1(5^{−1} × 2^{−1}) × 6^{−1}
=10−1×6−1=10^{−1}× 6^{−1}
=(10×6)−1= (10 × 6)^{−1} [∵am×bm=(ab)m] [∵a^m × b^m = (ab)^m]
=(60)−1=160= (60)^{−1} = \frac{1}{60} [∵a−m=1am][∵ a^{-m} = \frac{1}{a^m}]