I have this equation and I've been stuck with it for a couple of hours.
$\lfloor\log_2x\rfloor + 1 = \lceil\log_2(x+1)\rceil$
I've tried using this ceiling property:
$\lceil x\rceil = n \Leftarrow\Rightarrow x\le n\lt x+1$
Which gives me:
$\log_2(x+1)\le\lfloor\log_2x\rfloor + 1 \lt \log_2(x+1)+1$
But then, if I try to use the equivalent property for the floor function or try with the integer floor(x) + fractional part I get confused and I don't know how to proceed.
Thanks for your help!