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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

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

$\text { Let }:$

$x = 10$

$x + ∆ x = 10 . 02$

$\text { Then },$

$∆ x = 0 . 02$

$\text { For }x = ,$

$y = \log_e 10 = 2 . 3026$

$\text { Let }:$

$dx = ∆ x = 0 . 02$

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

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

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

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

$\Rightarrow ∆ y = 0 . 002$

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) (with solutions)
Chapter 14: Differentials, Errors and Approximations
Q: 9.09 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the Loge 10.02, It Being Given that Loge10 = 2.3026 . concept: Approximations. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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