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Question
Find the approximate values of : loge(101), given that loge10 = 2.3026.
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Solution
Let f(x) - logex.
Then f'(x) = `(1)/x`
Take a = 100 and h = 1. then
f(a) = f(100)
= loge100
= 2 loge10
= 2 x 2.3026
= 4.6052
f'(a) = f'(100)
= `(1)/(100)`
= 0.01
The formula for a approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ loge101 = f(101)
= f(100 + 1)
≑ f(100) + 1.f'(100)
≑ 4.6052 + 1 x 0.01
= 4.6152
loge(101) ≑ 4.6152.
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