---
title: "N-R method: x log10 x = 1.2"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:newton-raphson-x-log-x"
source: https://www.bctengineer.com/q/semester_4/numerical_method/newton-raphson-x-log-x
site: BCT Engineering
---

# N-R method: x log10 x = 1.2

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2075 Baisakh, 6 marks
- Video: [Find the root of the equation X.logX - 1.2 correct up to five decimal places ? #BSc_Maths #MSc_Maths](https://www.youtube.com/watch?v=cwIRsV8AzFY) — Maths Magic By Gaurav Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/newton-raphson-x-log-x
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Find a real root of the equation x log10(x) = 1.2 by N-R method correct up to 4 decimal places.

## Short answer

Let f(x)=x log10(x)-1.2, f'(x)=log10(x)+0.4343. Starting from x0=2.5, Newton-Raphson iteration x(n+1)=x(n)-f/f' gives root 2.7406 after 2 iterations.

## Answer

**Newton-Raphson formula:** x(n+1) = x(n) − f(x(n)) / f′(x(n)).

#### Step 1: Function and derivative
```text
f(x)  = x·log10(x) − 1.2
f'(x) = log10(x) + x·(1/(x·ln 10)) = log10(x) + 0.4343
```

#### Step 2: Initial guess
f(2) = 2(0.3010) − 1.2 = −0.5979 < 0 and f(3) = 3(0.4771) − 1.2 = 0.2314 > 0. The root lies in (2, 3), nearer to 3. Take x0 = 2.5.

#### Step 3: Iterations
| n | x(n) | f(x(n)) | f'(x(n)) | x(n+1) |
|---|---|---|---|---|
| 0 | 2.5000 | -0.2051 | 0.8322 | 2.7465 |
| 1 | 2.7465 | 0.0051 | 0.8731 | 2.7406 |
| 2 | 2.7406 | 0.0000 | 0.8721 | 2.7406 |

Successive values agree to four decimals (|x(n+1) − x(n)| < 0.00005).

**Answer: root ≈ 2.7406**

## Check yourself

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

1. f(x) for x log10 x = 1.2 is:
   A. x log10 x - 1.2
   B. x log10 x + 1.2
   C. log10 x - 1.2
   D. x ln x
   Answer: A. Move 1.2 to the left side. (Key point: Function)

2. f'(x) is:
   A. log10(x) + 1
   B. log10(x) + 0.4343
   C. 1/x
   D. log10(x) - 0.4343
   Answer: B. d/dx(x log10 x) = log10 x + x*(1/(x ln10)). (Key point: Derivative)

3. Root lies in (2, 3) because:
   A. f(2) > 0
   B. f(3) = 0
   C. f(2) < 0 and f(3) > 0
   D. f'(2) = 0
   Answer: C. The sign change brackets the root. (Key point: Initial guess)

4. The root to four decimals is:
   A. 2.5000
   B. 2.7465
   C. 3.0000
   D. 2.7406
   Answer: D. Iterations converge to 2.7406. (Key point: Result)

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