Ivars Wrote:3_log(a^b) =3_log(a[3]b)= 3_log(a)*3_log(b) = 3_log(a)[2]3_log(b)
This is where it starts to break down, because this is not a normal logarithm. Normal logarithms satisfy \( \log(a^b) = b\log(a) \)
Ivars Wrote:3_exp(a*b) =3_exp(a[2]b) = (3_exp(a))^(3_exp(b)) = 3_exp(a)[3]3_exp(b)
Since this is not the definition of logarithms, we should really separate out the 'log' notation from the equations:
\( \log(a^b) = \log(a) b \)
\( f(a^b) = f(a) f(b) \)
Using these equations we can evaluate at the following points:
If \( a=b^{1/b} \):
\( f(b) = f(b^{1/b}) f(b) \)
\( 1 = f(b^{1/b}) \)
then we find that f(a) is a constant.
If a=1:
\( f(1) = f(1) f(b) \)
then f(b) is a constant.
If b=1:
\( f(a) = f(a) f(1) \)
\( 1 = f(1) \)
then we find the point f(1)=1.
If b=0:
\( f(1) = f(a) f(0) \)
\( 1 = f(a) f(0) \)
which means f(a) is a constant.
So, not only can we show that f(x) is a constant, but we can show it in multiple ways, which is a very bad sign. I think this is proof that no such function can exist (or the only solution to all of these equations is f(x)=1).
Andrew Robbins