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

[thharrimw]

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

 

[mgb_phys]

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

 

[thharrimw]

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

but how are they based on calculus?

 

[mgb_phys]

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

 

[d_leet]

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

From 1 to a, not 0.

 

[mgb_phys]

do'h sorry

 

[lurflurf]

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

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.

 

[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

 

[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.

 

[thharrimw]

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.

is e a ratio?

 

[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.

 

[thharrimw]

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

 

[mgb_phys]

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

 

[HallsofIvy]

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

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

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

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