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Fit curve y = a b^x (decay data)

Short answer

Take log10 of y = ab^x to get Y = A + Bx. Least squares gives B = -0.2563 and A = 1.0835, so a = 10^A = 12.1199 and b = 10^B = 0.5543. The fitted curve is y = 12.1199(0.5543)^x.

Fit the following set of data to a curve of the form y = ab^x.

x1.01.52.02.53.03.54.0
y8.25.23.12.51.71.61.4

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.

xyY = log yx²xY
18.20.91381.00000.9138
1.55.20.71602.25001.0740
23.10.49144.00000.9827
2.52.50.39796.25000.9949
31.70.23049.00000.6913
3.51.60.204112.25000.7144
41.40.146116.00000.5845
Σx = 17.53.099850.75005.9557

Normal equations

ΣY = n·A + B·Σx      →  3.0998 = 7A + 17.5000B
ΣxY = A·Σx + B·Σx²   →  5.9557 = 17.5000A + 50.7500B

Solving: B = (nΣxY − ΣxΣY)/(nΣx² − (Σx)²) = -0.2563, A = (ΣY − BΣx)/n = 1.0835.

a = 10ᴬ = 12.1199, b = 10ᴮ = 0.5543.

Answer: y = 12.1199 · (0.5543)ˣ (decay curve, b < 1; at x = 2 it gives 3.72 against 3.1).

✓ Answer checked by Claude Opus 5.5 (high effort)

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