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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