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