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Solution for Using Differential, Find the Approximate Value of the Loge 10.02, It Being Given that Loge10 = 2.3026 . - CBSE (Science) Class 12 - Mathematics

Question

Using differential, find the approximate value of the loge 10.02, it being given that loge10 = 2.3026 ?

Solution

$\text { Consider the function } y = f\left( x \right) = \log_e x .$

$\text { Let }:$

$x = 10$

$x + ∆ x = 10 . 02$

$\text { Then },$

$∆ x = 0 . 02$

$\text { For }x = ,$

$y = \log_e 10 = 2 . 3026$

$\text { Let }:$

$dx = ∆ x = 0 . 02$

$\text { Now }, y = \log_e x$

$\Rightarrow \frac{dy}{dx} = \frac{1}{x}$

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 10} = \frac{1}{10}$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{10} \times 0 . 02 = 0 . 002$

$\Rightarrow ∆ y = 0 . 002$

$\therefore \log_e 10 . 02 = y + ∆ y = 2 . 3046$

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Solution Using Differential, Find the Approximate Value of the Loge 10.02, It Being Given that Loge10 = 2.3026 . Concept: Approximations.
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