Question 18

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waecmaths question: 

Given that $\frac{{{5}^{n+3}}}{{{25}^{2n-3}}}={{5}^{0}}$ , find n

Option A: 

n  = 1

Option B: 

n  = 2

Option C: 

n  = 3

Option D: 

n  = 5

waecmaths solution: 

$\begin{align}  & \frac{{{5}^{n+3}}}{{{25}^{2n-3}}}={{5}^{0}} \\ & \frac{{{5}^{n+3}}}{{{5}^{2(2n-3)}}}={{5}^{0}} \\ & {{5}^{n+3-2(2n-3)}}={{5}^{0}} \\ & {{5}^{n+3-4n+6}}={{5}^{0}} \\ & {{5}^{-3n+9}}={{5}^{0}} \\ & -3n+9=0 \\ & n=3 \\\end{align}$

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