N-R method: x log10 x = 1.2

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.

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

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

f(x)  = x·log10(x) − 1.2
f'(x) = log10(x) + x·(1/(x·ln 10)) = log10(x) + 0.4343

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.

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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