Question

The number of solutions of the equation$4^{x} -3^{x+\frac{1}{2}} -2^{2x-1, x \in R}$ is

Answer 1

$4^{x} -3^{x+\frac{1}{2}} =3^{x+\frac{1}{2}} -2^{2x -1}$
$\Rightarrow 4^{x} + 2^{2x -1} =3^{x+\frac{1}{2}} +3^{x- \frac{1}{2}}$
$\Rightarrow 2^{2x}\left(1+\frac{1}{2}\right)= 3^{x} \left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)$
$\Rightarrow 2^{2x} \left(\frac{3}{2}\right) =3^{x} \left(\frac{4}{\sqrt{3}}\right)$
$2^{2x -3} = 3^{x \frac{3}{2}}$
$\Rightarrow 2^{2x -3} =3^{\frac{2x -3}{2}}$
$2^{2x -3} = \left(\sqrt{3}\right)^{2x -3}$
which holds if $2x -3 =0\Rightarrow x =\frac{3}{2}$
The solution is $\left\{\frac{3}{2}\right\}$