---
title: "Fit y = a e^(bx)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:fit-y-aebx"
source: https://www.bctengineer.com/q/semester_4/numerical_method/fit-y-aebx
site: BCT Engineering
---

# Fit y = a e^(bx)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2069 Bhadra, 8 marks
- Video: [Curve fitting method || least square method || fitting of the exponential curve Y= ae^bx || #snme](https://www.youtube.com/watch?v=rI_kLs6hSSM) — Dev prit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/fit-y-aebx
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Fit the following set of data to a curve of the form y = a e^(bx) from the following observation by least square method.
| x | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| y | 5.5 | 6.5 | 9.4 | 15.2 | 30.6 | 49.8 |

## Short answer

Linearize via ln y = ln a + bx. Least squares on the transformed data gives A=1.0301, B=0.4613, so a=2.8015, b=0.4613. The curve is y = 2.8015 e^(0.4613x).

## Answer

Take natural logarithms: y = a·e^(bx) → ln y = ln a + b·x. Put Y = ln y, A = ln a, so Y = A + bx is a straight line.

| x | y | Y = ln y | x² | xY |
|---|---|---|---|---|
| 1 | 5.5 | 1.7047 | 1.0000 | 1.7047 |
| 2 | 6.5 | 1.8718 | 4.0000 | 3.7436 |
| 3 | 9.4 | 2.2407 | 9.0000 | 6.7221 |
| 4 | 15.2 | 2.7213 | 16.0000 | 10.8852 |
| 5 | 30.6 | 3.4210 | 25.0000 | 17.1050 |
| 6 | 49.8 | 3.9080 | 36.0000 | 23.4481 |
| Σx = 21 |  | 15.8676 | 91.0000 | 63.6088 |

#### Normal equations (b = B)

```text
ΣY = n·A + B·Σx      →  15.8676 = 6A + 21.0000B
ΣxY = A·Σx + B·Σx²   →  63.6088 = 21.0000A + 91.0000B
```

Solving: B = (nΣxY − ΣxΣY)/(nΣx² − (Σx)²) = 0.4613,  A = (ΣY − BΣx)/n = 1.0301.

a = e^A = 2.8015, b = 0.4613.

**Answer: y = 2.8015 e^(0.4613x)**  (at x = 6 it gives 44.60 against 49.8).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is y = a e^(bx) linearised?
   A. log y = a + b/x
   B. y = ln a + bx
   C. ln y = ln a + bx
   D. ln y = a ln x
   Answer: C. Natural log gives a straight line in x. (Key point: Logs)

2. How is a recovered from A?
   A. a = e^A
   B. a = A
   C. a = ln A
   D. a = 10^A
   Answer: A. A = ln a, so a = e^A = 2.8015. (Key point: a)

3. What is b in the final answer?
   A. ln B
   B. The slope B = 0.4613
   C. e^B
   D. 1/B
   Answer: B. Slope of ln y against x is b itself. (Key point: b)

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