waecmaths question:
Given that $t={{2}^{-x}}$ find ${{2}^{x+1}}$ in terms of t
Option A:
$\frac{2}{t}$
Option B:
$\frac{t}{2}$
Option C:
$\frac{1}{2t}$
Option D:
2t
waecmaths solution:
$\begin{align} & \text{Given }{{2}^{-x}}=y \\ & {{2}^{x+1}}={{2}^{x}}\times 2=\frac{1}{\tfrac{1}{{{2}^{x}}}}\times 2=\frac{1}{{{2}^{-x}}}\times 2=\frac{1}{y}\times 2=\frac{2}{y} \\ & \therefore {{2}^{x+1}}=\frac{2}{y} \\\end{align}$
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