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# Solution for Using Differential, Find the Approximate Value of the Loge 4.04, It Being Given that Log104 = 0.6021 and Log10e = 0.4343 ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the loge 4.04, it being given that log104 = 0.6021 and log10e = 0.4343 ?

#### Solution

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

$\text { Let }:$

$x = 4$

$x + ∆ x = 4 . 04$

$\text { Then },$

$∆ x = 0 . 04$

$\text { For } x = 4,$

$y = \log_e 4 = \frac{\log_{10} 4}{\log_{10} e} = \frac{0 . 6021}{0 . 4343} = 1 . 386368$

$\text { Let }:$

$dx = ∆ x = 0 . 04$

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

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

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 4} = \frac{1}{4}$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{4} \times 0 . 04 = 0 . 01$

$\Rightarrow ∆ y = 0 . 01$

$\therefore \log_e 4 . 04 = y + ∆ y = 1 . 396368$

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Solution Using Differential, Find the Approximate Value of the Loge 4.04, It Being Given that Log104 = 0.6021 and Log10e = 0.4343 ? Concept: Approximations.
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