Let g$($n$)$g$($n$)$g$($n$)$ be the number of times the statement x$=$x$+1$x $=$ x $+1$x$=$x$+1$ is executed in the following pseudocode:

x $=0$ i $=1$ while $($i $$ n$)\{$ i $=2*$ i x $=$ x $+1\}$

Order all of the following sentences in the most natural way so that they form a proof for the following statement:

g$($n$)=($log$2$n$)$g$($n$)=\backslash$Theta$(\backslash$log$\_2$ n$)$g$($n$)=($log$2$n$)$

Choose from these sentences:

This means that the number of times the statement i$=2$ii $=2*$ ii$=2$i is executed is equal to g$($n$)=$k$=$log$2$n$$g$($n$)=$ k $=\backslash$lceil $\backslash$log$\_2$ n $\backslash$rceilg$($n$)=$k$=$log$2$n$.$

Secondly, we have $$log$2$n$$log$2$n$\backslash$lceil $\backslash$log$\_2$ n $\backslash$rceil $\backslash$geq $\backslash$log$\_2$ n$$log$2$n$$log$2$n$,$ which implies $$log$2$n$=$O$($log$2$n$)\backslash$lceil $\backslash$log$\_2$ n $\backslash$rceil $=\backslash$mathcal$\{$O$\}(\backslash$log$\_2$ n$)$log$2$n$=$O$($log$2$n$).$

For each positive integer iii, if the statement i$=2$i$+$ii $=2^$i $+$ ii$=2$i$+$i is executed iii times, then iii becomes $2$i$2^$i$2$i$.$

Therefore, if kkk is the integer such that $2$k$1$