22 Oct 05
Originally posted by elopawnthe last one doesn't need a calculator, 2 *10^1,000,000
Write the following in Scientific Notation:
3^667 * 2^2 = ? * 10^?
2^10^6 = ? * 10^?
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.
22 Oct 05
1 edit
Originally posted by sonhouseOh wow I totally misread the first one. I thought it said 3*667.
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.
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.
23 Oct 05
Originally posted by AThousandYoung2^10 is 1028??? I was always told its 1024. They cheated me....took so much money....and taught me wrong. :'(
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.
- Joined
- 26 Apr 03
- Moves
- 26771
24 Oct 05
2 edits
Originally posted by elopawnThis seems a bit like it might be old O level maths homework about logs.
Write the following in Scientific Notation:
3^667 * 2^2 = ? * 10^?
2^10^6 = ? * 10^?
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
24 Oct 05
Originally posted by iamatigerI don't understand how you made this jump:
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
log(y) = 301029.9957
log(y) - log(9.9015) = 301029
- Joined
- 26 Apr 03
- Moves
- 26771
24 Oct 05
Originally posted by AThousandYoungAs per the technique in the first example
I don't understand how you made this jump:
[b]log(y) = 301029.9957
log(y) - log(9.9015) = 301029[/b]
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))