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

Question

empty · Answer

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

amistre64 · Answer

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)

amistre64 · Answer

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

empty · Answer

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.

anonymous · Answer

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.

amistre64 · Answer

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

amistre64 · Answer

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

empty · Answer

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.