Solving a exponential floor flunction equation I am trying to solve this question but can't simplify this further.
Question:
$$ 2^{\lfloor log_2{(x)} + \frac{1}{2} \rfloor} = 2^{\lfloor log_2(x-2^{\lfloor{ log_2{(\frac{x}{2})} + \frac{1}{2}}\rfloor}) + \frac{1}{2}\rfloor}  + 2^{\lfloor log_2{(\frac{x}{2})+\frac{1}{2}}\rfloor}$$
My first step
$$
= 2^{\lfloor{log_2{(x - 2^{\lfloor{log_2{(x)} + \frac{1}{2}-1}\rfloor})} + \frac{1}{2} }\rfloor} + 2^{\lfloor{log_2{(x) + \frac{1}{2} - 1}}\rfloor} $$
as : $log_2(x) + \frac{1}{2} = \lfloor{log_2(x) + \frac{1}{2}}\rfloor + r $
Then I rewrite $\lfloor{log_2(x) + \frac{1}{2}}\rfloor = log_2(x) + \frac{1}{2} - r $
So : $ 2^{\lfloor{log_2{(x)} + \frac{1}{2}}\rfloor} = 2^{log_2{(x)} + \frac{1}{2} -r} = x \cdot 2^{v}$ where $v = \frac{1}{2} - r$
So that
$$ 2^{\lfloor log_2{(x)} + \frac{1}{2} \rfloor} =  2^{\lfloor{log_2{(x - x\cdot2^{v-1} )} + \frac{1}{2}}\rfloor} + x \cdot 2^{v-1} $$
$$ = 2^{\lfloor{log_2{(x(1 - 2^{v-1}) )} + \frac{1}{2}}\rfloor} + x \cdot 2^{v-1}$$
$$ = 2^{\lfloor{log_2{(x)}+ \frac{1}{2}  + log_2{(1-2^{v-1})} }\rfloor} + x \cdot 2^{v-1} $$
I'm stuck here. Can anyone help?
 A: ${first}... \\ $
$\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} =\mathrm{2}.\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]}  \\ $
$. \\ $
${so}... \\ $
$. \\ $
$\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} −\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]}  \\ $
$. \\ $
${then}... \\ $
$. \\ $
${k}=\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} =\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \left({x}−{k}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]}  \\ $
$\frac{{x}}{\mathrm{2}}={x}−{k}\Rightarrow\frac{{x}}{\mathrm{2}}={k}=\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]}  \\ $
${for}:\:\left\{\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right\}\:\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $
$\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\left[\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right]\Rightarrow{x}=\mathrm{2}^{{n}} ,\:{n}\in{Z} \\ $
${for}:\:\left\{\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right\}\:\in\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right) \\ $
$\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\left[\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right]+\mathrm{1}\Rightarrow\left({no}\:{solution}\right) \\ $
$. \\ $
${so}\:{the}\:{answer}:\:{x}=\mathrm{2}^{{n}} ,\:{n}\in{Z} \\ $
