waecmaths question:
If $2{{\log }_{x}}(3\tfrac{3}{8})=6$ find the value of x
Option A:
$\tfrac{3}{2}$
Option B:
$\tfrac{4}{3}$
Option C:
$\tfrac{2}{3}$
Option D:
$\tfrac {1}{2}$
waecmaths solution:
$\begin{align} & 2{{\log }_{x}}(3\tfrac{3}{8})=6 \\ & {{\log }_{x}}(\tfrac{27}{8})=3 \\ & {{x}^{3}}=\frac{27}{8} \\ & {{x}^{3}}={{\left( \frac{3}{2} \right)}^{3}} \\ & x=\frac{3}{2} \\\end{align}$
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