Question

Find the approximate value of log₁₀ (1016), given that log₁₀e = 0⋅4343.

Answer 1

Let f(x) = log₁₀x = `(log_ex)/(log_e10)`

= (log₁₀e)(log_ex)

= (0.4343) log x

∴ f'(x) = `0.4343/x`

x = 1016 = 1000 + 16 = a + h

Here, a = 1000 and h = 16

f(a) = f(1000)

= log₁₀(1000)

= log₁₀(10)³

= 3log₁₀ 10   ...[∵ log10 mⁿ = n log₁₀ m]

= 3

f'(a) = f'(1000) = `0.4343/1000` = 0.0004343

f(a + h) ≈ f(a) + hf'(a)

log₁₀(1016) ≈ 3 + 16(0.0004343)

≈ 3 + 0.0069488

log₁₀(1016) ≈ 3.006949