---
title: "Fit y = a b^x and find y(7)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:fit-y-abx-y7"
source: https://www.bctengineer.com/q/semester_4/numerical_method/fit-y-abx-y7
site: BCT Engineering
---

# Fit y = a b^x and find y(7)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2070 Bhadra, 8 marks
- 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-abx-y7
- 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 b^x. Also evaluate y(7).
| x | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|
| y | 16.0 | 11.1 | 8.7 | 6.4 | 4.7 | 2.6 |

## Short answer

Take log: Y=log y. Normal equations yield A=1.3669, B=−0.0743, so a=23.2761, b=0.8428. Curve: y=23.2761·0.8428ˣ. y(7)≈7.031.

## Answer

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 |
|---|---|---|---|---|
| 2 | 16 | 1.2041 | 4.0000 | 2.4082 |
| 4 | 11.1 | 1.0453 | 16.0000 | 4.1813 |
| 6 | 8.7 | 0.9395 | 36.0000 | 5.6371 |
| 8 | 6.4 | 0.8062 | 64.0000 | 6.4494 |
| 10 | 4.7 | 0.6721 | 100.0000 | 6.7210 |
| 12 | 2.6 | 0.4150 | 144.0000 | 4.9797 |
| Σx = 42 |  | 5.0822 | 364.0000 | 30.3767 |

#### Normal equations

```text
ΣY = n·A + B·Σx      →  5.0822 = 6A + 42.0000B
ΣxY = A·Σx + B·Σx²   →  30.3767 = 42.0000A + 364.0000B
```

Solving: B = (nΣxY − ΣxΣY)/(nΣx² − (Σx)²) = -0.0743,  A = (ΣY − BΣx)/n = 1.3669.

a = 10ᴬ = 23.2761, b = 10ᴮ = 0.8428.

**Answer: y = 23.2761 · (0.8428)ˣ**

#### y(7)

```text
y(7) = 23.2761 × (0.8428)⁷ = 7.0313
```

**y(7) ≈ 7.031**

## Check yourself

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

1. How is y = a b^x fitted?
   A. Use log y = log a + x log b
   B. Use 1/y
   C. Use sqrt(y)
   D. Use y = a + bx
   Answer: A. Fit a straight line to Y = log y against x. (Key point: Linearisation)

2. How are a and b obtained?
   A. a = B, b = A
   B. Equal to A and B
   C. Logs of A and B
   D. Antilogs of A and B
   Answer: D. a = 10^A, b = 10^B. (Key point: Constants)

3. How is y(7) obtained?
   A. Substitute x = 7 in y = 23.2761 (0.8428)^x
   B. Average y(6) and y(8)
   C. Substitute in log y
   D. Use Lagrange
   Answer: A. Evaluate the fitted curve at x = 7, giving about 7.03. (Key point: Prediction)

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