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

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:

Using these equations we can evaluate at the following points:

If :

then we find that f(a) is a constant.

If a=1:

then f(b) is a constant.

If b=1:

then we find the point f(1)=1.

If b=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