Question

Simplify.

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

Answer 1

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

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

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

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

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

\(=\frac{ 625\ t^4}{2}\)

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

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

\(= 3^{−5−(−5)} × 2^{−5−(−5)} × 5^{−5−(−7)+3  }\)  [Since, \(a^m × a^n = a^{m + n}\) and \(\frac{a^m}{a^n} = a^{m − n}\)]
\(= 3^0 × 2^0 × 5^5\)
\(= 1 × 1 × 5^5\)     \([∵a^0=1]\)
\(= 5^5\)