Question
Simplify: $\frac{(9)^{3}\times27\times\text{t}^4}{(3)^{-2}\times(3)^{4}\times\text{t}^{2}}$
✓
Answer
$\frac{(9)^{3}\times27\times\text{t}^4}{(3)^{-2}\times(3)^{4}\times\text{t}^{2}}$
$=\frac{(3^2)^{3}\times(3)^3\times\text{t}^4}{(3)^{-2}\times(3)^{4}\times\text{t}^{2}}$
$[\because3\times3=3^2=9]$
$\frac{(3)^{6}\times(3)^3\times\text{t}^4}{(3)^{-2}\times(3)^{4}\times\text{t}^{2}}$
$=(3)^{6}\times(3)^3\times(3)^2\times(3)^{-4}\times\text{t}^4\times\text{t}^{-2}$
$\big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\big]$
$=(3)^{11-4}\times\text{t}^{4-2}=(3)^7\times\text{t}^2$
$[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m}+\text{n}}]$
$=\frac{(3^2)^{3}\times(3)^3\times\text{t}^4}{(3)^{-2}\times(3)^{4}\times\text{t}^{2}}$
$[\because3\times3=3^2=9]$
$\frac{(3)^{6}\times(3)^3\times\text{t}^4}{(3)^{-2}\times(3)^{4}\times\text{t}^{2}}$
$=(3)^{6}\times(3)^3\times(3)^2\times(3)^{-4}\times\text{t}^4\times\text{t}^{-2}$
$\big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\big]$
$=(3)^{11-4}\times\text{t}^{4-2}=(3)^7\times\text{t}^2$
$[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m}+\text{n}}]$
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