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Question

Mathematics Question on Laws of Exponents

Simplify.

  1. (25×t−4)(5−3×10×t−8)(t≠0)\frac{(25×t^{−4})}{(5^{−3}×10×t^{−8})} (t ≠ 0)
  2. (3−5×10−5×125)(5−7×6−5)\frac{ (3^{−5}×10^{−5}×125)}{(5^{−7}×6^{−5})}
Answer

(i) (25×t−4)(5−3×10×t−8)\frac{(25 × t^{−4})}{(5^{−3} × 10 × t^{−8})}

=(52×t−4)(5−3×5×2×t−8)= \frac{(5^2 × t^{−4})}{(5^{−3} × 5 × 2 × t^{−8} )}

=(52×t−4)(5−3+1×2×t−8)= \frac{(5^2 × t^{−4})}{(5^{−3 + 1} × 2 × t^{−8}) } [Since, am × an = am + n]

=(52×t−4)(5−2×2×t−8)= \frac{(5^2 × t^{−4})}{(5^{−2} × 2 × t^{−8})}

=(52−(−2)×t−4−(−8))2= \frac{(5^{2−(−2)} × t^{−4−(−8)})}{2} [Since, aman=am−n\frac{a^m}{a^n} = a^{m − n}]
=(54×t−4+8)2= \frac{(5^4 × t^{−4 + 8})}{2}

=625 t42=\frac{ 625\ t^4}{2}

(ii) (3−5×10−5×125)(5−7×6−5)\frac{(3^{−5}×10^{−5}×125)}{(5^{−7}× 6^{−5})}

=(3−5×(2×5)−5×53)(5−7×(2×3)−5)= \frac{(3^{−5} × (2 × 5)^{−5} × 5^3)}{(5^{−7}× (2 × 3)^{-5})}

=3−5−(−5)×2−5−(−5)×5−5−(−7)+3= 3^{−5−(−5)} × 2^{−5−(−5)} × 5^{−5−(−7)+3 } [Since, am×an=am+na^m × a^n = a^{m + n} and aman=am−n\frac{a^m}{a^n} = a^{m − n}]
=30×20×55= 3^0 × 2^0 × 5^5
=1×1×55= 1 × 1 × 5^5 [∵a0=1][∵a^0=1]
=55= 5^5