Logarithm Problem, Need Help

[ryanshep]

I got this as an Extra Credit problem and well... I really need this Extra Credit so here it is....This is hard to write with the computer but Ill try.



What is the Sum?



1/log(base)2 100!+1/log(base)3 100!+1/log(base)4 100!..... +1/log(base)100 100!



That's one divided by log base2 one hundred factorial then log base 3 etc etc all the way up to 100 anyone know the answer and how to get it? Thanks

*edit I need this answer asap its due tomorrow!

[ryanshep]

anyone?

[MayAMonkeyBeYourPinata]

parenthesise and then factorial

[ryanshep]

yeah but my stupid calculator cant factorial that high, i think its some sort of trick

[MayAMonkeyBeYourPinata]

If the numbers are continiously decreasing.

So that each time you add less and less, you can just do it up to say 10, because after that it only has a small effect.

[ryanshep]

Gracias

[Miscue]

You know, I was about to go to sleep but this problem interested me so I solved it for you.

The answer is 1

How:

The terms you are adding up look like: 1/LOGx(100!) where x goes from 2 to 100
BTW... my notation intends for LOGx(y) to mean the base x log of y

1. Note the identity: LOGa(b) = 1/LOGb(a).

1/LOGx(100!) = LOG100!(x) (Refer to identity #1)

Ok, so each term will look like: LOG100!(x) where x goes from 2 to 100. So, it's going to be LOG100!(2) + LOG100!(3) + ... + LOG100!(100)

2. Note the identity: LOGb(xy) = LOGb(x) + LOGb(y)
LOG100!(2) + LOG100!(3) + ... + LOG100!(100) = LOG100!(2*3*...*100) (Refer to identity #2)

LOG100!(2*3*...*100) = LOG100!(100!)

3. Note the identity: LOGb(b) = 1

LOG100!(100!) = 1 (Refer to identity #3)

[Automaggot68]

Miscue, that was so good, You owned me before I even opened this thread.
DAMN.
Good Job.

You know, I was about to go to sleep but this problem interested me so I solved it for you.

The answer is 1

How:

The terms you are adding up look like: 1/LOGx(100!) where x goes from 2 to 100
BTW... my notation intends for LOGx(y) to mean the base x log of y

1. Note the identity: LOGa(b) = 1/LOGb(a).

1/LOGx(100!) = LOG100!(x) (Refer to identity #1)

Ok, so each term will look like: LOG100!(x) where x goes from 2 to 100. So, it's going to be LOG100!(2) + LOG100!(3) + ... + LOG100!(100)

2. Note the identity: LOGb(xy) = LOGb(x) + LOGb(y)
LOG100!(2) + LOG100!(3) + ... + LOG100!(100) = LOG100!(2*3*...*100) (Refer to identity #2)

LOG100!(2*3*...*100) = LOG100!(100!)

3. Note the identity: LOGb(b) = 1

LOG100!(100!) = 1 (Refer to identity #3)

[Miscue]

Miscue, that was so good, You owned me before I even opened this thread.
DAMN.
Good Job.

When you guys get to doing these by hand:

http://www.oreilly.com/catalog/masteralgoc/chapter/ch13.html

Let me know...

Loads of fun doing curve fitting with a pencil and paper.