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

1. KingGeorge Group Title

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

2. johnnydiamond08 Group Title

3 or 2^3?

3. KingGeorge Group Title

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 Group Title

So it would be 4 then?

5. KingGeorge Group Title

Looks perfect to me.

6. johnnydiamond08 Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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

11. KingGeorge Group Title

Looks perfect.

12. johnnydiamond08 Group Title

Thanks again!

13. KingGeorge Group Title

You're welcome.