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Find the approximate values of : loge(9.01), given that log 3 = 1.0986.

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Question

Find the approximate values of : loge(9.01), given that log 3 = 1.0986.

Sum
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Solution

Let f(x) = logex

∴ f'(x) = `(1)/x`

x = 9.01 = 9 + 0.01 = a + h

Here, a = 9 and h = 0.01 

f(a) = f(9) = loge9
= loge32

= 2 loge 3

= 2 × 1.0986

= 2.1972

f'(a) = f'(9) = `1/9` = 0.1111

f(a + h) ≑ f(a) + h.f'(a)

∴ loge (9.01) ≑ 2.1972 + (0.01) (0.1111)

≑ 2.1972 + 0.001111

∴ loge (9.01) ≑ 2.198311

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Chapter 2: Applications of Derivatives - Exercise 2.2 [Page 75]

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