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Posers and Puzzles

Posers and Puzzles

  1. 22 Oct '05 07:20
    Write the following in Scientific Notation:

    3^667 * 2^2 = ? * 10^?

    2^10^6 = ? * 10^?
  2. Subscriber AThousandYoung
    It's about respect
    22 Oct '05 09:20
    Originally posted by elopawn
    Write the following in Scientific Notation:

    3^667 * 2^2 = ? * 10^?

    2^10^6 = ? * 10^?
    8*10^3

    1*10^18
  3. Subscriber sonhouse
    Fast and Curious
    22 Oct '05 20:44
    Originally posted by elopawn
    Write the following in Scientific Notation:

    3^667 * 2^2 = ? * 10^?

    2^10^6 = ? * 10^?
    the last one doesn't need a calculator, 2 *10^1,000,000
    but most calcs wouldn't be able to do that.
    in engineering notation, it would be 2E1,000,000.
    The first one is 3^667 * 4, my calc won't do that, maybe my HP48
    can, right now, can't find the dang thing. none of the regular
    ones do three digit exponents.
  4. Subscriber AThousandYoung
    It's about respect
    22 Oct '05 21:06 / 1 edit
    Originally posted by sonhouse
    the last one doesn't need a calculator, 2 *10^1,000,000
    but most calcs wouldn't be able to do that.
    in engineering notation, it would be 2E1,000,000.
    The first one is 3^667 * 4, my calc won't do that, maybe my HP48
    can, right now, can't find the dang thing. none of the regular
    ones do three digit exponents.
    Oh wow I totally misread the first one. I thought it said 3*667.

    The second one you are mistaken. 2 to any power is not necessarily going to be 2 to some power of 10. 2^10 is 1028. Then you need to raise that to the 6th power which I did via calculator.

    3^667 * 2^2 = 12(3^666)
    = 12((3^2)^333)
    = 12(9^333)
    = 12((9^3)^111)
    = 12(729^111)
    = 12((729^3)^37)
    = 12(387420489^37)

    This will take some entering stuff into the calculator which I am not in the mood to do, but it's doable with the calculator on the computer.
  5. Standard member gaurav2711
    walking...
    23 Oct '05 06:24
    Originally posted by AThousandYoung
    Oh wow I totally misread the first one. I thought it said 3*667.

    The second one you are mistaken. 2 to any power is not necessarily going to be 2 to some power of 10. 2^10 is 1028. Then you need to raise that to the 6th power which I did via calculator.

    2^10 is 1028??? I was always told its 1024. They cheated me....took so much money....and taught me wrong. :'(
  6. Subscriber AThousandYoung
    It's about respect
    24 Oct '05 06:34
    Originally posted by gaurav2711
    2^10 is 1028??? I was always told its 1024. They cheated me....took so much money....and taught me wrong. :'(
    Whatever. It's not 2*10^n.
  7. 24 Oct '05 16:05 / 2 edits
    Originally posted by elopawn
    Write the following in Scientific Notation:

    3^667 * 2^2 = ? * 10^?

    2^10^6 = ? * 10^?
    This seems a bit like it might be old O level maths homework about logs.

    x = 3^667 * 2 ^ 2

    take logs of both sides:
    log x = 667 log 3 + 2 log 2

    using excel (or log tables) for the logs
    log x = 318.8419369
    log x - 0.8419369 = 318

    using excel (or alog tables) to get 10^0.8416369
    log x - log 6.949267029 = 318

    alogs of both sides
    x/6.94926709 = 10^318
    x = 6.94926709 * 10 ^ 318

    -------------------------------

    similarly

    y = 2^(10^6)
    log(y) = 10^6.log(2)
    log(y) = 301029.9957
    log(y) - log(9.9015) = 301029

    y = 9.9015*10^301029
  8. Subscriber AThousandYoung
    It's about respect
    24 Oct '05 19:35
    Originally posted by iamatiger
    This seems a bit like it might be old O level maths homework about logs.

    x = 3^667 * 2 ^ 2

    take logs of both sides:
    log x = 667 log 3 + 2 log 2

    using excel (or log tables) for the logs
    log x = 318.8419369
    log x - 0.8419369 = 318

    using excel (or alog tables) to get 10^0.8416369
    log x - log 6.949267029 = 318

    alogs of both sides
    x/6.949267 ...[text shortened]... og(y) = 10^6.log(2)
    log(y) = 301029.9957
    log(y) - log(9.9015) = 301029

    y = 9.9015*10^301029
    I don't understand how you made this jump:

    log(y) = 301029.9957
    log(y) - log(9.9015) = 301029
  9. 24 Oct '05 20:07
    Originally posted by AThousandYoung
    I don't understand how you made this jump:

    [b]log(y) = 301029.9957
    log(y) - log(9.9015) = 301029
    [/b]
    As per the technique in the first example

    where floor() means the number rounded to the nearest integer towards zero, and rem() means the remaider of the number after floor:

    if log(y) = a

    then log(y) = floor(a) + rem(a)

    log(y) - rem(a) = floor(a)

    log(y) - log(10^(rem(a)) = floor(a)

    y/(10^(rem(a)) = 10^(floor(a))

    y = 10^(rem(a)) * 10^(floor(a))

    so the 9.9015 in your question was simply 10^0.9957 = 10^(rem(a))

    But I suppose I could have done it more simply:

    if log(y) = a

    log(y) = floor(a) + rem(a)

    y = 10^(floor(a) + rem(a))

    and because j^(k + L) = j^k * j^L:

    y = 10^(rem(a)) * 10^(floor(a))