# Scientific Notation.

elopawn
Posers and Puzzles 22 Oct '05 07:20
1. 22 Oct '05 07:20
Write the following in Scientific Notation:

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

2^10^6 = ? * 10^?
2. AThousandYoung
All My Soldiers...
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. 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. AThousandYoung
All My Soldiers...
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. 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. AThousandYoung
All My Soldiers...
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 16:052 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. AThousandYoung
All My Soldiers...
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))