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johnnydiamond08

  • 3 years ago

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

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  1. KingGeorge
    • 3 years ago
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    First up, you need to find what \(\log_2(8)\) is. Can you tell me what that is?

  2. johnnydiamond08
    • 3 years ago
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    3 or 2^3?

  3. KingGeorge
    • 3 years ago
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    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
    • 3 years ago
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    So it would be 4 then?

  5. KingGeorge
    • 3 years ago
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    Looks perfect to me.

  6. johnnydiamond08
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    So following the same equation but making X 123,456 the answer would be 17.

  11. KingGeorge
    • 3 years ago
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    Looks perfect.

  12. johnnydiamond08
    • 3 years ago
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    Thanks again!

  13. KingGeorge
    • 3 years ago
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    You're welcome.

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