weights in multiplication I understand I have an equation of two fractions $0 < A, B < 1$.
$$R = \dfrac{A+B}2,$$ where $0 < R, A, B <1$. I understand how to bias towards A or B by inserting weights k1,k2 to the equation, like
$$R = \dfrac{k_1A + k_2B}2,$$ where $k_1+k_2=1$.
how do I bias towards a variable in a multiplication scenario, i.e., $R = AB$ ? I sense it should be something like $R = (A+k_1)(B+k_2)$ but I cannot find the boundary conditions and relationship for $k_1,k_2$. Would appreciate your help!
By the way my range for A, B is 1/7 -> 1, with a step of 1/7th.
So A*B should be 1/49 (or as close as possible)
Thanks,
Andreas 
 A: Because you started your question by asking about a sum which is really an arithmetic mean, people who responded have tended to interpret your "multiplication" question as a question about the geometric mean,
because the geometric mean has the same relationship to multiplication that the arithmetic mean has to addition.
For a weighted geometric mean, you would choose $k_1$ and $k_2$ such that 
$k_1+k_2=1,$ $0 \leq k_1 \leq 1,$ and $0 \leq k_2 \leq 1,$ and then the weighted geometric mean is
$$ R=A^{k_1}B^{k_2}.$$
Note that I wrote more conditions on $k_1$ and $k_2$ than you really need; in effect, I wrote four inequalities (two for $k_1,$ two for $k_2$) where only two are actually needed. For example, $0 \leq k_1$ and $0 \leq k_2$ would have been enough.
Alternatively, with the same conditions on $k_1$ and $k_2$ you could write
$$ R=A^{1 - k_1}B^{1 - k_2},$$
and this also would give a weighted geometric mean of the same kind, because the conditions imply that
$(1-k_1)+(1-k_2)=1,$ $0 \leq (1-k_1) \leq 1,$ and $0 \leq (1-k_2) \leq 1.$
The fact that $0 < R, A, B \leq 1$ is not a problem. These formulas work for any positive numbers.
Also observe that when $k_1 = k_2 = \frac12,$ you get the ordinary geometric mean,
$$ R = A^{1/2}B^{1/2} = \sqrt{AB}.$$
If you actually want a weighted product, so that equal weighting would give you
$R = AB,$ then you can simply double the exponents in the geometric mean.
For example, you can set
$$ R=A^{2k_1}B^{2k_2},$$
which gives you $R = AB$ when $k_1 = k_2 = \frac12.$
If you would prefer not to write the $2$ in each exponent, you can change the conditions so that $k_1+k_2=2,$ $0 \leq k_1 \leq 2,$ and $0 \leq k_2 \leq 2,$
and write
$$ R=A^{k_1}B^{k_2},$$
which gives you $R = AB$ when $k_1 = k_2 = 1$ and otherwise gives you a product weighted toward whichever factor has the greater exponent.
A: From your first example, it looks like you want a formula for $R = f(k,A,B)$ where $\lim_{k \to 1} R = A$, and $\lim_{k \to 0} R = B$. The following is one possibility: $R = A^k B^{1-k}$, but I am not sure if this will behave as you want for any $0<k<1$.
Since R is the mean of A,B in your first example, perhaps you are looking for a weighted geometric mean?
$R = e^{k \ln(A)+(1-k)\ln(B)}$
