Question

How do I go about solving: 1+Floor(Log2(x)) where x = 8

Answer 1

First up, you need to find what \(\log_2(8)\) is. Can you tell me what that is?

Answer 2

3 or 2^3?

Answer 3

Right. It's 3. So now you have the equation \[1+\lfloor3\rfloor\]Can you finish it from here? Or do you need some help with the floor function?

Answer 4

So it would be 4 then?

Answer 5

Looks perfect to me.

Answer 6

Okay so just to make sure I know what I'm doing. All you do is find what X would be and then add one to it?

Answer 7

The floor function is a bit weird. For example, if you had \[1+\left\lfloor \log_2(9)\right\rfloor\]instead, you would simplify as follows.

Since \(\log_2(9)\approx 3.17\), you have\[1+\left\lfloor \log_2(9)\right\rfloor\\
1+\left\lfloor 3.17\right\rfloor \\
1+3 \\
4\]
In general, the floor function gets rid of any numbers to the right of the decimal place.

Answer 8

Ah alright. We're doing all this stuff by hand so my professor is making it pretty easy for us. Thank you KingGeorge, you've been a lot of help!

Answer 9

Formally, \[\left\lfloor x\right\rfloor=\text{max}\{m\in\mathbb{Z} |\;m<x\]In laymans terms, the greatest integer that is not greater than \(x\).

Answer 10

So following the same equation but making X 123,456 the answer would be 17.

Answer 11

Looks perfect.

Answer 12

Thanks again!

Answer 13

You're welcome.