- 22 Oct '05 20:44

the last one doesn't need a calculator, 2 *10^1,000,000*Originally posted by elopawn***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 21:06 / 1 edit

Oh wow I totally misread the first one. I thought it said 3*667.*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.

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 06:24

2^10 is 1028??? I was always told its 1024. They cheated me....took so much money....and taught me wrong. :'(*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.

- 24 Oct '05 16:05 / 2 edits

This seems a bit like it might be old O level maths homework about logs.*Originally posted by elopawn***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 19:35

I don't understand how you made this jump:*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

**log(y) = 301029.9957**

log(y) - log(9.9015) = 301029 - 24 Oct '05 20:07

As per the technique in the first example*Originally posted by AThousandYoung***I don't understand how you made this jump:**[/b]

[b]log(y) = 301029.9957

log(y) - log(9.9015) = 301029

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))