Question 7

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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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