Explanation

(b)
Given equation is
$4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$
$\Rightarrow 2^{2x} + 2^{2x - 1} = \ 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$
$\Rightarrow 2^{2x}\left( \ 1 + \frac{1}{2} \right) = 3^{x - \frac{1}{2}}(3 + 1)$
$\Rightarrow \ 2^{2x}.\frac{3}{2} = 3^{x - \frac{1}{2}}.4$
$\Rightarrow \ 2^{2x - 3} = 3^{x - \frac{3}{2}}$
Taking log on both sides, we get
$(2x - 3)\log{2 = \left( \ x - \frac{3}{2} \right)\log 3}$
$\Rightarrow 2x\ \log\ 2 - 3\log{2 = x\log{3 - \frac{3}{2}\log 3}}$
$\Rightarrow x\log{4 - x\log{3 = 3\log{2 - \frac{3}{2}\log 3}}}$
$\Rightarrow x\log{\left( \frac{4}{3} \right) = \log{8 - \log{3\sqrt{3}}}}$
$\Rightarrow \log{\left( \frac{4}{3} \right)^{x} = \log\frac{8}{3\sqrt{3}}}$
$\Rightarrow \ \left( \frac{4}{3} \right)^{x} = \frac{8}{3\sqrt{3}}$
$\Rightarrow \ \left( \frac{4}{3} \right)^{x} = \left( \frac{4}{3} \right)^{3/2}$
$\therefore x = \frac{3}{2}$