I can not follow you in this post.
Also your linear construction on sci.math.research is not completely clear to me.
Also if you say unique then you should specify under what conditions.
Specifically I dont know what you mean with the function "frac" in your post in sci.math.research.
Let me gather what I got so far. If we have the conditions
\( {}^1 b = b \) and \( {}^{x+1}b=b^{{}^x b} \) (1)
then to define \( g(x)={}^{x}b \) it suffices to define an initial function f on the interval \( (0,1] \) and
then derive the value on an interval \( (n,n+1\] \) via \( g(x)=\exp_b^{\circ n}(f(x-n)) \) (where \( \exp_b^{\circ n} \) is the n-times iterated \( \exp_b(x)=b^x \)).
For intervals \( (-n,-n+1] \) left of \( (0,1] \) one has similarly \( g(x)=\log_b^{\circ n}(f(x+n)) \).
If \( g(x) \) is continuous at 1 then
\( g(1)=b^{g(0)} \), \( g(1)=b^{b^{g(-1)}} \) and \( g(1)=b^{b^{b^{g(-2)}}} \) from which \( {}^0b=1 \), \( {}^{-1}b=0 \), \( {}^{-2}b=-\infty \) follows.
So it makes sense to define g on \( (-2,\infty) \).
Then somehow you wanted to define the initial function \( f : (0,1]\to(1,b] \) to be linear and from there stopped my understanding.
Especially I didnt see the problem you mentioned with \( b=e^e,10,1.5 \).
Edit: Now it seems to me that \( \text{frac}(x)=1+(b-1)*x \) is the initial linear function that maps \( (0,1]\to (1,b] \). Ok, but if I graph the resulting function g(x), I see angles at x=0 and x=1, for both \( b=e \) and \( b=e^{1/e} \).
Also your linear construction on sci.math.research is not completely clear to me.
Also if you say unique then you should specify under what conditions.
Specifically I dont know what you mean with the function "frac" in your post in sci.math.research.
Let me gather what I got so far. If we have the conditions
\( {}^1 b = b \) and \( {}^{x+1}b=b^{{}^x b} \) (1)
then to define \( g(x)={}^{x}b \) it suffices to define an initial function f on the interval \( (0,1] \) and
then derive the value on an interval \( (n,n+1\] \) via \( g(x)=\exp_b^{\circ n}(f(x-n)) \) (where \( \exp_b^{\circ n} \) is the n-times iterated \( \exp_b(x)=b^x \)).
For intervals \( (-n,-n+1] \) left of \( (0,1] \) one has similarly \( g(x)=\log_b^{\circ n}(f(x+n)) \).
If \( g(x) \) is continuous at 1 then
\( g(1)=b^{g(0)} \), \( g(1)=b^{b^{g(-1)}} \) and \( g(1)=b^{b^{b^{g(-2)}}} \) from which \( {}^0b=1 \), \( {}^{-1}b=0 \), \( {}^{-2}b=-\infty \) follows.
So it makes sense to define g on \( (-2,\infty) \).
Then somehow you wanted to define the initial function \( f : (0,1]\to(1,b] \) to be linear and from there stopped my understanding.
Especially I didnt see the problem you mentioned with \( b=e^e,10,1.5 \).
Edit: Now it seems to me that \( \text{frac}(x)=1+(b-1)*x \) is the initial linear function that maps \( (0,1]\to (1,b] \). Ok, but if I graph the resulting function g(x), I see angles at x=0 and x=1, for both \( b=e \) and \( b=e^{1/e} \).