anonymous
  • anonymous
How do I go about solving: 1+Floor(Log2(x)) where x = 8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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KingGeorge
  • KingGeorge
First up, you need to find what \(\log_2(8)\) is. Can you tell me what that is?
anonymous
  • anonymous
3 or 2^3?
KingGeorge
  • 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?

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anonymous
  • anonymous
So it would be 4 then?
KingGeorge
  • KingGeorge
Looks perfect to me.
anonymous
  • anonymous
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?
KingGeorge
  • 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.
anonymous
  • anonymous
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!
KingGeorge
  • KingGeorge
Formally, \[\left\lfloor x\right\rfloor=\text{max}\{m\in\mathbb{Z} |\;m
anonymous
  • anonymous
So following the same equation but making X 123,456 the answer would be 17.
KingGeorge
  • KingGeorge
Looks perfect.
anonymous
  • anonymous
Thanks again!
KingGeorge
  • KingGeorge
You're welcome.

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