Question
Simplify: $\frac{(3^{-2})^{2}\times(5^2)^{-3}\times(\text{t}^{-3})^2}{(3^{-2})^{5}\times(5^3)^{-2}\times(\text{t}^{-4})^3}$
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Answer
$\frac{(3^{-2})^{2}\times(5^2)^{-3}\times(\text{t}^{-3})^2}{(3^{-2})^{5}\times(5^3)^{-2}\times(\text{t}^{-4})^3}$
$=\frac{(3)^{-4}\times(5)^{-6}\times\text{t}^{-6}}{(3)^{-10}\times(5)^{-6}\times\text{(t)}^{-12}}$
$[\because(\text{a}^{\text{m}})^\text{n}=(\text{a})^\text{mn}]$
$=(3)^{-4}\times(3)^{10}\times(5)^{-6}\times(5)^{6}\times(\text{t})^{-6}\times(\text{t})^{12}$
$=(3)^{-4+12}\times(5)^{-6+6}\times(\text{t})^{-6+12}$
$\big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\big]$
$=(3)^6\times5^0\times(\text{t})^6=(3\text{t})^6$
$[\because\text{a}^0=1]$
$=\frac{(3)^{-4}\times(5)^{-6}\times\text{t}^{-6}}{(3)^{-10}\times(5)^{-6}\times\text{(t)}^{-12}}$
$[\because(\text{a}^{\text{m}})^\text{n}=(\text{a})^\text{mn}]$
$=(3)^{-4}\times(3)^{10}\times(5)^{-6}\times(5)^{6}\times(\text{t})^{-6}\times(\text{t})^{12}$
$=(3)^{-4+12}\times(5)^{-6+6}\times(\text{t})^{-6+12}$
$\big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\big]$
$=(3)^6\times5^0\times(\text{t})^6=(3\text{t})^6$
$[\because\text{a}^0=1]$
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