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  • one year ago

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

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  1. Empty
    • one year ago
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    \[a^{f(a)}=e\] \[a^{f(b)}=1\] for \(a \ne b\)

  2. amistre64
    • one year ago
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    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)

  3. amistre64
    • one year ago
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    can a=1? 1^k = e doesnt seem plausible to me, so a=1 is not an issue.

  4. Empty
    • one year ago
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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.

  5. anonymous
    • one year ago
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    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.

  6. amistre64
    • one year ago
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    \[\Huge a^{\frac{x-b}{(a-b)ln(a)}}\] \[\Huge a^{\frac{b-b}{(a-b)ln(a)}}=a^{f(b)\color{red}{=0}}=1\] \[\Huge a^{\frac{a-b}{(a-b)ln(a)}}=a^{f(a)\color{red}{=\frac{1}{ln(a)}}}=e\]

  7. amistre64
    • one year ago
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    log_a(e) change of bases to ln(e)/ln(a) hence the 1/ln(a)

  8. Empty
    • one year ago
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    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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