- 2068 Bhadra · 8 marks
Fit y = a b^x (growth data)
Short answer
Linearize to log y = log a + x log b. Solving the normal equations yields log a = -0.2811, log b = 0.3555, so a = 0.5235, b = 2.2674. The fitted curve is y = 0.5235(2.2674)^x.
Find a curve of the form y = ab^x that fits the following set of observations using least square method.
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | 1.2 | 2.5 | 6.25 | 15.75 | 28.65 |
Take logarithms (base 10): y = a·bˣ → log y = log a + x·log b. Put Y = log y, A = log a, B = log b, so Y = A + Bx is a straight line in x.
| x | y | Y = log y | x² | xY |
|---|---|---|---|---|
| 1 | 1.2 | 0.0792 | 1.0000 | 0.0792 |
| 2 | 2.5 | 0.3979 | 4.0000 | 0.7959 |
| 3 | 6.25 | 0.7959 | 9.0000 | 2.3876 |
| 4 | 15.75 | 1.1973 | 16.0000 | 4.7891 |
| 5 | 28.65 | 1.4571 | 25.0000 | 7.2856 |
| Σx = 15 | 3.9274 | 55.0000 | 15.3374 |
Normal equations
ΣY = n·A + B·Σx → 3.9274 = 5A + 15.0000B
ΣxY = A·Σx + B·Σx² → 15.3374 = 15.0000A + 55.0000B
Solving: B = (nΣxY − ΣxΣY)/(nΣx² − (Σx)²) = 0.3555, A = (ΣY − BΣx)/n = -0.2811.
a = 10ᴬ = 0.5235, b = 10ᴮ = 2.2674.
Answer: y = 0.5235 · (2.2674)ˣ (growth curve, b > 1; at x = 5 it gives 31.37 against 28.65).
✓ Answer checked by Claude Opus 5.5 (high effort)
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