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Provide solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (ix)

Answers (1)


Answer: 2.3046

Hint: Here we use

        \Delta y=f(x+\Delta x)-f(x)

Given: \log e 10.2=2.3026


Consider y=\operatorname{loge} x

Here x=10 \text { and } \Delta x=0.02

By Differentiating  w.r.t x,

So we get

\Delta y=\frac{d y}{d x} \Delta x

\Rightarrow \Delta y=\frac{d y}{d x} \Delta x=\frac{1}{10} \times 0.02

on further calculation

\begin{aligned} &\Delta y=\frac{0.02}{10}=0.002 \\\\ &\Rightarrow \Delta y=f(x+\Delta x)-f(x) \\\\ &0.002=\operatorname{loge}(10+0.02)-\operatorname{loge} 10 \end{aligned}

It can be written as

0.002=\operatorname{loge} 10.02=2.3026

We get,

\log 10.02=2.3026


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