I should really be able to do this on my own but I need help with the following problem.
F (x)= D
F (2x)= D + 3
F (4x) = D + 6
…
I need the inverse function G(D+A)=Bx so that if I know D and A I can calculate B.
What you wrote doesn’t really make sense.
In general you have
F(x * 2^N) = D + 3*N
but assuming D is a constant, how does the right hand side depend on x?
ETA: Maybe you shouldn’t have the X there at all. Does this fit with what you’re trying to do?
F (1)= D
F (2)= D + 3
F (4) = D + 6
…
F(2^N) = D + 3*N
Then in general, you’d have
F(X) = D + 3 * log[sub]2/sub
D is not constant
OK so what it sounds like you really want is a function F such that for all x,
F(2x)=F(x)+3
This is clearly impossible for x=0, since it would require that
F(0)=F(0)+3.
However, if we eliminate this point, and we set for X not equal to 0
F(x)=3log[sub]2/sub=3log(|x|)/log(2)
we find that
F(2x)= 3log(|2x|)/log(2) = 3(log(2)+log(|x|))/log(2) = 3+3*log(|x|)/log(2) = 3+F(x)
we have a solution.
PS: you can also add any constant to the above solution and it still works.
So the full set of solutions is of the form
F(x)=3*log[sub]2/sub+C for any constant C and x not equal to 0. Basically what ZenBeam said.
Thanks for the help guys!