Ok, so:
I suppose that the exponential function you mean is this one: y = ae^(bx).
To find a and b, we replace:
6400 = ae^(2b) (1 equation)
4096 = ae^(4b). (2 equation)
We can solve this system as this:
If we find a in the first equation:
a = (6400) / (e^(2b).
Now we replace this fact in equation 2:
4096 = (6400)/e^(2b)) * (e^(4b)).
Simplifying:
4096 = 6400 * e^(2b)
0.64 = e^(2b)
Now, we apply the natural logarithm to both sides:
ln (0.64) = ln(e^(2b)).
This is
ln (0.64) = 2b.
Then, b= ln(0.64)/2. which is approximately -0.22.
Now, we find a if we replace b in any equation:
6400 = ae^(2b)
6400 = ae^(2(-0.22))
6400 = ae^(-0.44))
a = 6400 / e^(-0.44) which is approximately 9937.3