Question

Simplify.(i)$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}},t\ne 0$, (ii)$\frac{3^{-5}\times {10}^{-5}\times 125}{5^{-7}\times 6^{-5}}$

Answer 1

(i) Given $\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}},t\ne 0$
$\Rightarrow \frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}=\frac{5^2\times t^{-4}}{5^{-3}\times 5\times 2\times t^{-8}}$
$=\frac{5^2\times t^{-4}}{5^{-3+1}\times 2\times t^{-8}}$ [Since, $a^m\times a^n=a^{m+n}$]
$=\frac{5^2\times t^{-4}}{5^{-2}\times 2\times t^{-8}}$
$=\frac{5^{2-(-2)}\times t^{-4-(-8)}}{2}$ [Since, $\frac{a^m}{a^n}=a^{m-n}$]
$=\frac{5^{2+2}\times t^{-4+8}}{2}$
$=\frac{5^4\times t^4}{2}$
$\therefore \frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}=\frac{625t^4}{2}$

(ii) Given $\frac{3^{-5}\times {10}^{-5}\times 125}{5^{-7}\times 6^{-5}}=\frac{3^{-5}\times {(2\times 5)}^{-5}\times 5^3}{5^{-7}\times {(2\times 3)}^{-5}}$
$=\frac{3^{-5}\times 2^{-5}\times 5^{-5}\times 5^3}{5^{-7}\times 2^{-5}\times 3^{-5}}$ [Since, $(a\times b{)}^n=a^n\times b^n$
$=3^{-5+5}\times 2^{-5+5}\times 5^{-5+3+7}$
$=3^0\times 2^0\times 5^5$
$=1\times 1\times 5^5$ [Since, $a^0=1$]
$\therefore \frac{3^{-5}\times {10}^{-5}\times 125}{5^{-7}\times 6^{-5}}=5^5$