Proof e x </msup> = exp &#x2061;<!-- ⁡ --> ( x ) ? Define

mistergoneo7 2022-07-03 Answered
Proof e x = exp ( x )?
Define
ln ( x ) = 1 x 1 t
Assume I have proven that ln x is one-to-one and therefore has an inverse exp ( x )
Define e as:
ln e = 1
Now, if you have no other notion of exponentials, or logarithms, how could define what e x means and show that its the inverse of ln x?
You are allowed to assume the logarithmic product and quotient property.
Thanks for the help.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Hayley Mccarthy
Answered 2022-07-04 Author has 19 answers
Writing f for ln and g for its inverse, you can show easily that g is infinitely differentiable and that g ( n ) ( 0 ) = 1. This gives you the Taylor series g ( x ) = n = 0 x n / n !. After that, everything follows from the classical analysis of g that is performed in every elementary real variables text (see Rudin's Real & Complex Analysis, for example).
As I recall, the introductory "Chapter 0" of that text is a marvel of succinct mathematics that fully constructs the exponential function from scratch. It's really a pleasure to read and I'm always awed at his insight every time I read it.

We have step-by-step solutions for your answer!

logiski9s
Answered 2022-07-05 Author has 1 answers
Full disclosure: this is essentially a rewrite of MPW's answer.
Defining e x as k = 0 x k k ! makes sense, in a way it's the most fundamental/general definition because it can be applied to any system for which addition, multiplication and scaling are defined. Reals, complex numbers, quaternions, matrices, etc.
Now, from calculus we have this result:
d d x [ f 1 ( x ) ] = 1 f ( f 1 ( x ) )
By fundamental theorem of calculus we obtain ln ( x ) as 1 x , hence:
d d x [ ln 1 ( x ) ] = ln 1 ( x )
It follows (formally by induction) that the nth derivative of ln 1 ( x ) is ln 1 ( x ) and hence the nth derivative at x = 0 is ln 1 ( 0 ) = 1
Thus we get the Taylor series:
ln 1 ( x ) = k = 0 x k k ! = e x

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more