## johnnydiamond08 How do I go about solving: 1+Floor(Log2(x)) where x = 8 one year ago one year ago

1. KingGeorge

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

2. johnnydiamond08

3 or 2^3?

3. KingGeorge

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?

4. johnnydiamond08

So it would be 4 then?

5. KingGeorge

Looks perfect to me.

6. johnnydiamond08

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?

7. KingGeorge

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.

8. johnnydiamond08

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!

9. KingGeorge

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 $$x$$.

10. johnnydiamond08

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

11. KingGeorge

Looks perfect.

12. johnnydiamond08

Thanks again!

13. KingGeorge

You're welcome.