Identities between exponential and logarithmic functions?

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The discussion centers on the identities between exponential and logarithmic functions, particularly focusing on their inverse relationship, such as e^x and ln(x). The original poster seeks a way to substitute 1/ln(x) with an exponential function to solve the integral ∫sin(x)/ln(x)dx. However, it is noted that this integral cannot be solved using elementary functions, and finding a series solution is also challenging. Suggestions of exploring complex numbers are made, but it is concluded that this approach would not yield a solution either. The conversation emphasizes the limitations of current methods in solving the integral.
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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