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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 (Science) Class 12 - Mathematics

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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 for question: Using Differential, Find the Approximate Value of the Loge 4.04, It Being Given that Log104 = 0.6021 and Log10e = 0.4343 ? concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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