waecmaths question:
Simplify $\frac{{{3}^{n-1}}\times {{27}^{n+1}}}{{{81}^{n}}}$
Option A:
32n
Option B:
9
Option C:
3n
Option D:
3n + 1
waecmaths solution:
$\begin{align} & \frac{{{3}^{n-1}}\times {{27}^{n+1}}}{{{81}^{n}}}=\frac{{{3}^{n-1}}\times {{3}^{3(n+1)}}}{{{3}^{4n}}} \\ & \frac{{{3}^{n-1}}\times {{27}^{n+1}}}{{{81}^{n}}}=\frac{{{3}^{n-1}}\times {{3}^{3n+3}}}{{{3}^{4n}}} \\ & \frac{{{3}^{n-1}}\times {{27}^{n+1}}}{{{81}^{n}}}=\frac{{{3}^{n-1+3n+3}}}{{{3}^{4n}}}=\frac{{{3}^{4n+2}}}{{{3}^{4n}}} \\ & \frac{{{3}^{n-1}}\times {{27}^{n+1}}}{{{81}^{n}}}={{3}^{4n+2-4n}}={{3}^{2}}=9 \\\end{align}$
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