Identities between exponential and logarithmic functions?

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Discussion Overview

The discussion revolves around the relationships and identities between exponential and logarithmic functions, particularly in the context of solving integrals. Participants explore various mathematical properties and potential substitutions involving these functions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Some participants note that the exponential function and logarithm are inverses, with basic relationships such as \( x = \ln(f) = \exp(g) \) and \( \ln(\exp(x)) = \exp(\ln(x)) \).
  • One participant expresses uncertainty about what specific identities are being sought, suggesting that any other identities could be derived from the inverse relationship.
  • Another participant seeks a way to substitute \( \frac{1}{\ln(x)} \) with an exponential function to help solve the integral \( \int \frac{\sin(x)}{\ln(x)} \, dx \), but finds that their proposed substitution \( \frac{1}{\ln(x)} = \log_x e \) does not eliminate the logarithmic function.
  • There is mention that the integral in question is not solvable using elementary functions and that finding a series solution is also challenging.
  • One participant suggests exploring the complex plane for potential solutions, referencing a substitution involving \( \sin(x) \), but expresses uncertainty about the implications of working with complex numbers.
  • Another participant counters that using complex numbers would not change the solvability of the integral, citing Mathematica's algorithm for determining solutions.

Areas of Agreement / Disagreement

Participants express various viewpoints on the relationships between exponential and logarithmic functions, but there is no consensus on specific identities beyond the inverse relationship. The discussion regarding the integral remains unresolved, with differing opinions on the applicability of complex analysis.

Contextual Notes

Participants mention limitations regarding the solvability of the integral using elementary functions and the challenges associated with finding a series solution. There is also uncertainty about the impact of complex analysis on the problem.

romsofia
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Hello! I was wondering if they're any identities between exponential and logarithmic functions? Maybe identities isn't the right word, but what I'm talking about is something like euler's formula. Other than than {e^x} and {ln(x)} are inverses, if that counts.

Any help is very much appreciative!
 
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Maybe identities isn't the right word,...
Let f=exp(x) and g=ln(x)
Of course there are basic relationships :
x = ln(f) = exp(g)
ln(exp(x)) = exp(ln(x))
 
JJacquelin said:
Let f=exp(x) and g=ln(x)
Of course there are basic relationships :
x = ln(f) = exp(g)
ln(exp(x)) = exp(ln(x))

AKA inverses, as I said other than that. Thanks for the help though.
 
Well, since these inverses are the definition of the logarithm, it follows that any other identity can be derived from those. So I'm not sure what kind of identities you want?? It would help us if you told us what you are looking for and why...
 
micromass said:
Well, since these inverses are the definition of the logarithm, it follows that any other identity can be derived from those. So I'm not sure what kind of identities you want?? It would help us if you told us what you are looking for and why...
I'm looking for a way to substitute {1/ln(x)} with an exponential function to help solve an integral (the integral, I've asked about it here before, is \int sin(x)/ln(x)\,dx sorry, don't know how to do fractions within integrals in latex!). Only substitution I can think of is {1/ln(x)=\log_x e}. However, that won't help since it's still a logarithmic function!
 
romsofia said:
I'm looking for a way to substitute {1/ln(x)} with an exponential function to help solve an integral (the integral, I've asked about it here before, is \int sin(x)/ln(x)\,dx sorry, don't know how to do fractions within integrals in latex!). Only substitution I can think of is {1/ln(x)=\log_x e}. However, that won't help since it's still a logarithmic function!

The integral (right click to see how I did the fraction)

\int{\frac{\sin(x)}{ln(x)}dx}

is not solvable using elementary functions, so you won't be able to solve it. The best thing to do is to find a series solution, which is also not easy.
 
micromass said:
The integral (right click to see how I did the fraction)

\int{\frac{\sin(x)}{ln(x)}dx}

is not solvable using elementary functions, so you won't be able to solve it. The best thing to do is to find a series solution, which is also not easy.
Thanks for showing on how to do fractions in integrals
Well, I was thinking if we were to go to the complex plane for the problem, then maybe we could solve it? Since we can make the substitution {sin(x)=1/2i(e^{-ix}-e^{ix})} However, I haven't had much exposure to complex numbers so I'm not really sure if it would make a difference if we were to work in the complex plane! However, I'm still lost on any exponential functions that we would be able to substitute for {1/ln(x)}
 
romsofia said:
Thanks for showing on how to do fractions in integrals
Well, I was thinking if we were to go to the complex plane for the problem, then maybe we could solve it? Since we can make the substitution {sin(x)=1/2i(e^{-ix}-e^{ix})} However, I haven't had much exposure to complex numbers so I'm not really sure if it would make a difference if we were to work in the complex plane! However, I'm still lost on any exponential functions that we would be able to substitute for {1/ln(x)}

Going to the complex numbers wouldn't make any difference. Mathematica uses an algorithm that decides 100% if there is a solution, and if it says that there isn't, then there isn't (even when working with complex numbers).
 
micromass said:
Going to the complex numbers wouldn't make any difference. Mathematica uses an algorithm that decides 100% if there is a solution, and if it says that there isn't, then there isn't (even when working with complex numbers).


Ahhh, okay thanks for all the help once again micromass :D!
 

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