A community for students.
Here's the question you clicked on:
 0 viewing
Empty
 one year ago
Is there some function f(x) that satisfies this:
Empty
 one year ago
Is there some function f(x) that satisfies this:

This Question is Closed

Empty
 one year ago
Best ResponseYou've already chosen the best response.0\[a^{f(a)}=e\] \[a^{f(b)}=1\] for \(a \ne b\)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0f(b) = 0 f(a) = 1/ln(a) we have 2 points to play with, can we make a line? m = 1/[(ab) ln(a)] f(x) = m(xb) +f(b)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0can a=1? 1^k = e doesnt seem plausible to me, so a=1 is not an issue.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ahh 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
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0\[\Huge a^{\frac{xb}{(ab)ln(a)}}\] \[\Huge a^{\frac{bb}{(ab)ln(a)}}=a^{f(b)\color{red}{=0}}=1\] \[\Huge a^{\frac{ab}{(ab)ln(a)}}=a^{f(a)\color{red}{=\frac{1}{ln(a)}}}=e\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0log_a(e) change of bases to ln(e)/ln(a) hence the 1/ln(a)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.