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

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

    2^10^6 = ? * 10^?
  2. SubscriberAThousandYoung
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    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. Subscribersonhouse
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    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. SubscriberAThousandYoung
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    22 Oct '05 21:061 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 membergaurav2711
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    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. SubscriberAThousandYoung
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    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. Joined
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    24 Oct '05 16:052 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. SubscriberAThousandYoung
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    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. Joined
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    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))
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