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# Solution for Using Differential, Find the Approximate Value of the Log10 10.1, It Being Given that Log10e = 0.4343 ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the  log10 10.1, it being given that log10e = 0.4343 ?

#### Solution

$\text { Consider the function y } = f\left( x \right) = \log_{10} x .$

$\text { Let }:$

$x = 10$

$x + ∆ x = 10 . 1$

$\text { Then },$

$∆ x = 0 . 1$

$\text { For } x = ,$

$y = \log_{10} 10 = 1$

$\text { Let }:$

$dx = ∆ x = 0 . 1$

$\text { Now,} y = \log_{10} x = \frac{\log_e x}{\log_e 10}$

$\Rightarrow \frac{dy}{dx} = \frac{1}{2 . 3025x}$

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 10} = 0 . 04343$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = 0 . 04343 \times 0 . 1 = 0 . 004343$

$\Rightarrow ∆ y = 0 . 004343$

$\therefore \log_{10} 10 . 1 = y + ∆ y = 1 . 004343$

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: 9.1 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the Log10 10.1, It Being Given that Log10e = 0.4343 ? concept: Approximations. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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