WAEC Further Mathematics 2015 Paper

$4(2^{x^2}) = 8^{x} \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}$

$\Rightarrow 2^{2 + x^{2}} = 2^{3x}$

Comparing bases, we have

$2 + x^{2} = 3x \Rightarrow x^{2} - 3x + 2 = 0$

$x^{2} - 2x - x + 2 = 0$

$x(x - 2) - 1(x - 2) = 0$

$(x - 1) = 0$ or $(x - 2) = 0$

$x = \;\text{1 or 2}\;$