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johnnydiamond08
How do I go about solving: 1+Floor(Log2(x)) where x = 8
First up, you need to find what \(\log_2(8)\) is. Can you tell me what that is?
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?
So it would be 4 then?
Looks perfect to me.
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?
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.
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!
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 \(x\).
So following the same equation but making X 123,456 the answer would be 17.