Logaritmic equation with floor brackets $ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor=\log_{10}x$

Question

Solve equation

$ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor= \log_{10}x$

I dont understand solution explained in the book.

$ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor= \left \lfloor \frac{x}{2}- \left \lfloor \frac{x}{2} \right \rfloor -\frac{1}{2} \right \rfloor=\left \lfloor \left \{ \frac{x}{2} \right \}-\frac{1}{2} \right \rfloor$

I dont understand why is

$\left \lfloor \frac{x}{2}- \left \lfloor \frac{x}{2} \right \rfloor -\frac{1}{2} \right \rfloor=\left \lfloor \left \{ \frac{x}{2} \right \}-\frac{1}{2} \right \rfloor$

Answer 1

$\{x\}$, the fractional part of x, is defined as $\{x\} = x - \lfloor x\rfloor$. (For example, $\{3.14\} = 0.14$, $\{-0.6\} = 0.4$, $\{1\} = 0$, etc. Thus, $\boxed{\frac{x}{2}-\left\lfloor\frac{x}{2}\right\rfloor = \bigg\{\frac{x}{2}\bigg\}.}$