## Sunday, November 8, 2015

### Floor function system

Solve the following equation in the real number system:

$\left\lfloor{\log_2 x}\right\rfloor+\left\lfloor{\log_4 x}\right\rfloor=3$

Since $\log_2x>\log_4x$, we can formulate two possibilities to make the sum of the LHS as $3$, i.e. $\left\lfloor{\log_2 x}\right\rfloor+\left\lfloor{\log_4 x}\right\rfloor=3+0\stackrel{\text{or}}=2+1$, but since $\log_2x=3$ gives $x=8$ and $\log_4x=0$ gives $x=1$, we know $x\lt 8$ and $\log_2x=2$ gives $x=4$ and $\log_4x=1$ gives $x=4$, so we can conclude the range of values of $x$ that satisfies $\left\lfloor{\log_2 x}\right\rfloor+\left\lfloor{\log_4 x}\right\rfloor=3$ is $[4, 8)$.