CBSE (Commerce) Class 12CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

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

Login
Create free account


      Forgot password?

Question

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

Solution

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

\[\text { Let }: \]

\[ x = 4 \]

\[x + ∆ x = 4 . 04\]

\[\text { Then }, \]

\[ ∆ x = 0 . 04\]

\[\text { For } x = 4, \]

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

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 04\]

\[\text { Now }, y = \log_e x\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x}\]

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

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

\[ \Rightarrow ∆ y = 0 . 01\]

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

  Is there an error in this question or solution?

APPEARS IN

Video TutorialsVIEW ALL [2]

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.
S
View in app×