## Empty one year ago Is there some function f(x) that satisfies this:

1. Empty

$a^{f(a)}=e$ $a^{f(b)}=1$ for $$a \ne b$$

2. amistre64

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

can a=1? 1^k = e doesnt seem plausible to me, so a=1 is not an issue.

4. Empty

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

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

$\Huge a^{\frac{x-b}{(a-b)ln(a)}}$ $\Huge a^{\frac{b-b}{(a-b)ln(a)}}=a^{f(b)\color{red}{=0}}=1$ $\Huge a^{\frac{a-b}{(a-b)ln(a)}}=a^{f(a)\color{red}{=\frac{1}{ln(a)}}}=e$

7. amistre64

log_a(e) change of bases to ln(e)/ln(a) hence the 1/ln(a)

8. Empty

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.