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

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KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2First up, you need to find what \(\log_2(8)\) is. Can you tell me what that is?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Right. 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So it would be 4 then?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Looks perfect to me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay 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?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2The 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah 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!

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Formally, \[\left\lfloor x\right\rfloor=\text{max}\{m\in\mathbb{Z} \;m<x\]In laymans terms, the greatest integer that is not greater than \(x\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So following the same equation but making X 123,456 the answer would be 17.
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