I'm not sure if this is better suited for an Excel forum, but here goes.
I will know the value of x, which can be anything between 1.00 and 71.32 (inclusive), and can have any number of decimal places to the right, such as 20.123456789.
An x value of EXACTLY 1.00 must return a value of EXACTLY 0.60
An x value of EXACTLY 71.32 must return a value of EXACTLY 1.00
The result values in the continuous range 0.60 to 1.00 must increase at a Decreasing rate. For example,
(1) An x value of around 2 should return a value greater than 0.60, call it z1
(2) An x value of around 70.32 should return a value less than 1.00, call it z2
Because I want a decreasing rate of increase, z1 minus 0.60 should be significantly greater than 1.00 minus z2
It seems like raising x to a power that is in the range 0 to 1 should provide the increase at a Decreasing rate, but nailing down the correct power that enforces both endpoints is where I'm stuck.
I haven't used logarithms in many years, so this is a struggle. Anyway, I put the following into Excel and I'm getting a #VALUE! message:
Cell A1 is a = 2/LOG(1783/25,5)
Cell A2 is b = 125/(1783*SQRT(1783))
Cell A3 is x = 20.123456789
Cell A4 is the result = A1*LN(A3/A2)
Please tell me where I'm screwing up before I move on to the Excel forum. Thanks much.
Let $var$ be the number that is currently $73.32$
$a=\dfrac{2}{5 \ln(var)}$
$b=(var)^{-3/2}$
so
a1 = 2/(5*LN(var))
a2 = (var)^(-3/2)
maybe what you'd want to do is make
a1=var
and shift a1 and a2 down one cell to become a2 and a3, then replace var with a1 in the formulas.
I thought I was done, but no. If I decided that the slope of the rate is too steep for the lower part of the range, how could I play around with different what-if slopes (rates) to find the optimum one (in my mind)? I can see that your 1783 is 25*71.32, and your 125 is 25*5, so I'm guessing that the number 5 and/or the -3/2 power is controlling the steepness. I will appreciate any further help I can get.