Adjust the data up curve φ(x) = α1e^(α2x) by the method of least squares:
Here's what I've done so far but I think it is wrong(and sorry for the bad english)
--x | 0 | 1 | 2 | 3 | 4 | 5 | 6
f(x)| 32 | 47 | 65 | 92 | 132 | 190 | 275
taking: φ(x) = α1e^(α2x), Z = ln(y) = ln(α1e^(α2x)) = ln(α1)+α2x = Θ(x)
and: Θ(x) = a1 + a2x, a1=ln(α1), a2=α2
--x | 0 | 1 | 2 | 3 | 4 | 5 | 6
Z=ln(y)| 3,465 | 3,850 | 4,174 | 4,522 | 4,883 | 5,247 | 5,617
Applying the least square method:
{ 6 a1 + Σ xk a2 = Σ Zk
{ Σ xk a1 + Σ xk² a2 = Σ z(xk).xk
=
{ 6 a1 + 21 a2 = 31,758
{ 21 a1 + 91 a2 = 105,233
a1=6,477 and a2=-0,3383
in the formula: α1 = e^a1=e^6,477=650,012
α2=a2=-0,3383
so... :
φ(x) = α1.e^(α2x) = 650,012.e^-0,3383x
I don't have the exacly answer, but the letter b) doesn't fit. Thank you for your time.
Hint
I checked all calculations and everything seems to be fine provided that $6$ becomes $7$. In the so-called normal equations, appears as a first coefficient the number of data points itself.
So, the equations I obtained are $$7{a_1}+21 {a_2}=31.7587$$ $$21 {a_1}+91 {a_2}=105.231$$ These are effectively the same as your (except the small error) and the solution is $$a_1=3.47031 \ \ \ a_2=0.355545$$ then $\alpha_1=e^{3.47031}=32.1469$ and the model is $$\phi(x)=32.1469\, e^{0.35545 x}$$ from which the computed data are $$\{32.1469,45.8722,65.4578,93.4056,133.286,190.193,271.398\}$$ which is indeed a very good fit.