Using differentials, find the approximate values of the following… - Vidyadip

Question

Using differentials, find the approximate values of the following:
$\log_\text{e}10.02$ it being given that $\log_\text{e}10=2.3026$

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Answer

Consider the function $\text{y}=\text{f} (\text{x})=\log_\text{e}\text{x}$ Let:
x = 10 $\text{x}+\triangle \text{x}=10.02$Then,
$\triangle\text{x} =0.02$ For x $\text{y}\log_\text{e} 10=2.3026$Let:
$\text{dx}=\triangle \text{x}=0.02$Now, $\text{y}=\log_\text {e}\text{x}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{\text {x}}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =10}=\frac{1}{10}$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {10}\times0.02=0.002$ $\Rightarrow\text{y} =0.002$ $\therefore\log_\text {e}10.02=\text{y}+\triangle\text{y} =2.3046$

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