WAEC 2008 · Paper 2 · Q4

Indices, logarithms & surds

(a)
Solve $2^{x + 1} - 4\left(2^{-x}\right) - 7 = 0$ .

Worked solution (try it first)

Let $y = 2^x$ .
Then $2^{x + 1} = 2y$ and $2^{-x} = \frac1y$ .
The equation becomes $2y - \frac4y - 7 = 0$ .
Multiply by $y$ : $2y^2 - 7y - 4 = 0$ .
Factorise: $(2y + 1)(y - 4) = 0$ , so $y = -\frac12$ or $y = 4$ .
$2^x$ is always positive, so reject $y = -\frac12$ .
$2^x = 4 = 2^2$ , so $x = 2$ .

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