Hi , I need help on the following:
Find a value of x such that :
$\displaystyle x = 4(\lceil log_2 (x)\rceil + 1) $
Thanks
The "ceiling" of $\displaystyle log_2(x)$, and so the right side, changes only when x is a power of two. If $\displaystyle x= 2^n$, $\displaystyle 4(log_2(x)+ 1)= 4(n+1)$. In order for that to be equal to $\displaystyle x= 2^n$ we must have $\displaystyle 2^n= 4(n+1)$, $\displaystyle 2^n/4= 2^{n-2}= n+1$.
You might be able to solve that algebraically using the Lambert W function but I would try a numerical method:
If n= 1, $\displaystyle 2^{-1}= 1/2\ne 1+ 1$. If n= 2 we have $\displaystyle 2^0= 1< 2+1$. If n= 3 we have $\displaystyle 2^1= 2< 3+ 1$. If n= 4 we have $\displaystyle 2^2= 4< 4+1$. If n= 5 we have $\displaystyle 2^3= 8> 5+ 1$. That tells us that there must be a solution to $\displaystyle 2^{n- 2}= n+ 1$ for n a number between n= 4 and n= 5. Okay, try half way between (just because that is simplest). If n= 4.5, [tex]2^{2.5}= 4\sqrt{2}[/quote] which is just slightly more than 4.5+ 1= 4.6. There must be a solution between n= 4 and 4.5 (and so x between $\displaystyle 2^4= 16$ and [tex]2^{4.5}= 22.67...