Let's start off with a simple function say $y = x$. Can it be written in terms of the natural logarithm? If so, are there any functions that cannot?
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Does $\ln y = \ln x$ count? – William Grannis Nov 14 '18 at 15:11
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1Could you elaborate the question? – Atharva Kathale Nov 14 '18 at 15:12
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Of course, this only works as long as both $x$ and $y$ are greater than $0$. – William Grannis Nov 14 '18 at 15:12
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@JohnNash That is not rigorous, as you assume x and y are both larger than 0 without expressing it explicitly. – William Grannis Nov 14 '18 at 15:21
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For any $x$ we have that
$$y=x \iff y=\ln e^x$$
and more in general
$$y=f(x) \iff y=\ln e^{f(x)}$$
Otherwise if we are interested in a $\log-\log$ identity
$$y=f(x) \implies \ln y = \ln (f(x))$$
is true only for $f(x)>0$.
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1Although this is cheating, as by log rules we get $y = f(x) \ln e $ – William Grannis Nov 14 '18 at 15:14
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There are plenty of functions which cannot. For example, consider Ackermann's Function. It can be rigorously shown that $A(x)$ is not expressible in any way save recursively. You could add an $\ln$ to the definition, but it would still not be expressible in closed form.
William Grannis
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I'm not sure what "it can be rigorously shown that $A(x)$ is not expressible in any way save recursively" means. It's true that the Ackermann function isn't primitive recursive, but I don't see why that implies what you've written. – Noah Schweber Nov 14 '18 at 15:21