Is there some function f(x) that satisfies this:

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Is there some function f(x) that satisfies this:

Mathematics
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\[a^{f(a)}=e\] \[a^{f(b)}=1\] for \(a \ne b\)
f(b) = 0 f(a) = 1/ln(a) we have 2 points to play with, can we make a line? m = 1/[(a-b) ln(a)] f(x) = m(x-b) +f(b)
can a=1? 1^k = e doesnt seem plausible to me, so a=1 is not an issue.

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Ahh well the main thing that they have to satisfy is that a and b are not the same whole number and that they can be any whole number not just 1. Kind of the crux of why this is bothering me haha.
I'm not sure there's anything that works besides the trivial function, \[f(x)=\begin{cases}\log_x e&\text{for }x=a,\quad a\in\mathbb{R}_+\backslash\{1\}\\0&\text{for }x\neq a\end{cases}\](where \(\mathbb{R}_+\) is the set of all positive reals) As amistre mentioned, \(x\neq1\) because that would suggest \(\log_1e=\frac{\ln e}{\ln1}\) exists, which it doesn't.
\[\Huge a^{\frac{x-b}{(a-b)ln(a)}}\] \[\Huge a^{\frac{b-b}{(a-b)ln(a)}}=a^{f(b)\color{red}{=0}}=1\] \[\Huge a^{\frac{a-b}{(a-b)ln(a)}}=a^{f(a)\color{red}{=\frac{1}{ln(a)}}}=e\]
log_a(e) change of bases to ln(e)/ln(a) hence the 1/ln(a)
Hmmm well the point of this is so that I can take any number, and do stuff like this: \[\log 200 = 3 \log 2 + 2 \log 5 \] Multiplying by f(2) or f(5) gives: \[f(2) \log 200 = 3\]\[f(5) \log 200 = 2\] Really the point is that it is essentially doing a dot product with a projection on a vector space to get that component \[\bar e_i \cdot \bar v = v_i\] Except the components are the exponents on the prime factorization.

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