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Differentials, Errors and Approximations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentials, Errors and Approximations5 Marks
Question
Using differentials, find the approximate values of the following:
$\log_\text{e}4.04,$ it being given that $\log_{{10}^{4}}=0.6021$ and $\log_{10}\text{e}=0.4343$
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Answer

Consider the function $\text{y}=\text{f}(\text{x})\log_\text {e}\text{x}.$Let:
x = 4 $\text{x}+\triangle \text{x}=4.04$Then,
$\triangle\text{x} =0.04$ For x = 4 $\text{y}=\log_\text {e}4=\frac{\log_{10}4}{\log_{10}\text {e}}=\frac{0.6021}{0.4343}=1.386368$ Let:
$\text{dx}=\triangle \text{x}=0.04$Now, $\text{y}=\log_\text {e}\text{x}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{\text {x}}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=4} =\frac{1}{4}$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {4}\times0.04=0.01$ $\Rightarrow\triangle \text{y}=0.01$ $\therefore\log_\text {e}4.04=\text{y}+\triangle\text{y} =1.39638$

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