Question

Assuming log₁₀ e = 0.4343, find an approximate value of Iog₁₀ 1003

Answer 1

Let f(x) = log ₁₀ x then

f'(x) = ${\displaystyle \frac{1}{x}}$ log₁₀ e(log₁₀ x = log₁₀ e log_e x)

f(x + Δx) – f(x) = f ‘(x) Δ

f(1003) – f(1000) = ${\displaystyle \frac{0.4344}{1000}\times 3}$

log₁₀ 1003 – log₁₀ 1000 = 0.0013029

log₁₀ 1003 = log₁₀ 10³ + 0.0013029

= 3 + 0.0013029

= 3.0013029

Approximate value of log₁₀ 1003 = 3.0013029