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discrete math functions problem?

Let f(n) = lg(n) be the log 2 function. Find f(2048).

4 Answers

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  • 1 decade ago
    Favorite Answer

    you are just looking for the value of log_2 (2048). Since 2^11 = 2048, then log_2 (2048) = 11

  • Pascal
    Lv 7
    1 decade ago

    lg 2048 = 11

    It would probably be a good idea for you to memorize the first few powers of 2:

    2^2=4

    2^3=8

    2^4=16

    2^5=32

    2^6=64

    2^7=128

    2^8=256

    2^9=512

    2^10=1024

    2^11=2048

    2^12=4096

    2^13=8192

    2^14=16,384

    2^15=32,768

    2^16=65,536

    2^17=131,072

    2^18=262,144

    2^19=524,288

    2^20=1,048,576

  • 1 decade ago

    That is easy because 2048 = 2^20

    And f(2048) = lg(2^20) = 20lg(2)=20(1) = 20

  • ?
    Lv 4
    5 years ago

    remember the fog(x) or f(g(x)) notation, dude? this is in certainty fixing for the fee of f(x) whilst x=g(x). think of of it this variety. permit f(x)=x+2. as quickly as we are asked to unravel for f(2), we replace 2 with x, subsequently-> f(2)=2+2=4. Composition of applications issues -> comparable technique. even however, fairly of numbers, we are comparing with expressions. a. f(x) = x +2 f(f(x)) = (x+2) + 2 changing x with f(x), or x+2 f(f(x)) = x + 4 b. f(x) = 3x f(f(x)) = 3(3x) f(f(x)) = 9x changing x with f(x) that's 3x Bro, i think of you have a typo in merchandise c. i'm hoping this permits.

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