Question

Simplify:
$\frac{25\times t^{-4}}{5^{-2}\times 10\times t^{-8}}\left(t\ne 0\right)$

Answer 1

$\frac{25\times t^{-1}}{5^{-2}\times 10\times t^{-1}}$

=$\frac{5^2\times t^{-1}}{5^{-2}\times 10\times t^{-1}}$

Since, $x^a\cdot x^b=x^{a+b}\&\frac{x^a}{x^b}=x^{a-b}$

=$\frac{5^2\times t^{-1}}{5^{-2}\times 5\times 2\times t^{-1}}$

=$\frac{5^2\times t^{-1}}{5^{-2+1}\times 2\times t^{-1}}$

=$\frac{5^2\times t^{-1}}{5^{-1}\times 2\times t^{-1}}$

=$\frac{5^2\times t^{-1}\times t^1\times 5^1}{2}$

=$\frac{5^2\times t^{-1}\times t^1\times 5^1}{2}$

=$\frac{5^3\times t^0}{2}$

=$\frac{5^3\times 1}{2}$

=$\frac{5^3}{2}$

=$\frac{125}{2}$