CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

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

Login
Create free account


      Forgot password?

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

Video TutorialsVIEW ALL [2]

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)
S
View in app×