Do Logarithms play any major part in being able to do calculus?

1. Feb 18, 2008

thharrimw

do Logarithms play any major part in being able to do calculus?

2. Feb 18, 2008

mgb_phys

More the other way around - the definition of ln() and 'e' are based on calculus.

3. Feb 18, 2008

thharrimw

but how are they based on calculus?

4. Feb 18, 2008

mgb_phys

The definition of ln(a) is the integral of 1/x from 0 to 'a'

5. Feb 18, 2008

d_leet

From 1 to a, not 0.

6. Feb 18, 2008

mgb_phys

do'h sorry

7. Feb 18, 2008

lurflurf

Your definition not everyones. It would be fair to say that many standard definitions use some concept of limit (i.e. use calculus) and many that do not use inequalities that are very limit like. Also we would like log to be continous, a calculus (or topology) concept.

8. Feb 19, 2008

thharrimw

ok so one of my teachers told me that the number e came from (1+1/n)^n as n Approaches infinity but now do you get 2.71828182846........ from (1+1/n)^n

9. Feb 19, 2008

mgb_phys

Try it! Just put in the first few terms.
To answer your original question, 'e' comes up in a few standard calculus solutuions.
And obviously knowing the rules about multiplying and adding exponents comes into a lot of calculus.

10. Feb 19, 2008

thharrimw

is e a ratio?

11. Feb 19, 2008

mgb_phys

No, it's an irrational number, that means that it can't be written as a/b.
Actually it's also a trancendental number - meaning it can't be written as any equation with a finite number of terms, just like pi.

Last edited: Feb 19, 2008
12. Feb 20, 2008

thharrimw

ok so the number e $$\approx$$ (1+1/n)^n

13. Feb 20, 2008

mgb_phys

I suppose a mathematician would say that e was exactly (1+/1n)^n for infiinite 'n'

14. Feb 20, 2008

HallsofIvy

Staff Emeritus
Not until you tell us what n is! Yes, for very large n, e is approximately that.

Not unless he/she were speaking very loosely. 'n' is never infinite. What is strictly true is that $e= \lim_{n\rightarrow \infty}(1+ 1/n)^n$.