Iterating powers of a number

Let’s analyze the loop below where we iterate over all powers of $2$

for (int i = 1; i <= N; i *= 2)
    x++;

• Iteration $1$: $i=1$
• Iteration $2$: $i=2$
• Iteration $3$: $i=4$
• …
• Iteration $x$: $i=2^{x-1}$

Let’s say the loop terminates after the $j^{th}$ iteration, i.e.,

$2^{j-1} <= N$

$j-1 <= logN$

$j <= logN + 1$

Hence, the loop runs $(logN + 1)$ times.

In Big-O notation, the time complexity is $O(logN)$ ...