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

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प्रश्न

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

बेरीज
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उत्तर

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

= (log10e)(logex)

= (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

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पाठ 2: Applications of Derivatives - Exercise 2.2 [पृष्ठ ७५]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 2 Applications of Derivatives
Exercise 2.2 | Q 5.3 | पृष्ठ ७५

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