WAEC 2021 · Paper 1 · Q10

If $16 \times 2^{(x + 1)} = 4^x \times 8^{(1 - x)}$ , find the value of $x$ .

Worked solution (try it first)

Write everything as a power of 2.
Left: $16 \times 2^{x + 1} = 2^4 \times 2^{x + 1}$
$= 2^{x + 5}$ .
Right: $4^x = 2^{2x}$ and $8^{1 - x} = 2^{3 - 3x}$ , so the right side is $2^{2x + 3 - 3x} = 2^{3 - x}$ .
Equate the powers: $x + 5 = 3 - x$ , so $2x = -2$ and $x = -1$ , option D.