waecmaths question:
Evaluate $\frac{{{\log }_{3}}9-{{\log }_{2}}8}{{{\log }_{3}}9}$
Option A:
$-\tfrac{1}{3}$
Option B:
$\tfrac{1}{2}$
Option C:
$\tfrac{1}{3}$
Option D:
$-\tfrac{1}{2}$
waecmaths solution:
$\begin{align} & \frac{{{\log }_{3}}9-{{\log }_{2}}8}{{{\log }_{3}}9}=\frac{{{\log }_{3}}{{3}^{2}}-{{\log }_{2}}{{2}^{3}}}{{{\log }_{3}}{{3}^{2}}} \\ & \frac{{{\log }_{3}}9-{{\log }_{2}}8}{{{\log }_{3}}9}=\frac{2{{\log }_{3}}3-3{{\log }_{2}}2}{2{{\log }_{3}}3} \\ & \frac{{{\log }_{3}}9-{{\log }_{2}}8}{{{\log }_{3}}9}=\frac{2-3}{2}\left| {{\log }_{a}}a \right.=1 \\\end{align}$
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