Find the approximate value of log_ 10 1005, given that log_ 10 e … - Vidyadip

Question

Find the approximate value of $log_{10} 1005$, given that $log_{10} e = 0.4343.$

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Answer

Let:
$\text{y}=\text{f}(\text{x})=\log_{10}\text{x}$
here,
x = 1000,
$\text{x}+\triangle\text{x}=1005$
$\Rightarrow\triangle\text{x}=5$
$\Rightarrow\text{dx}=\triangle\text{x}=5$
For x = 1000,
$\text{y}=\log_{10}1000=\log_{10}(10)^3=3$
Now, $\text{y}=\log_{10}\text{x}=\frac{\log_\text{e}\text{x}}{\log_\text{e}10}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{0.4343}{\text{x}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1000}=\frac{0.4343}{1000}=0.0004343$
$\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=0.0004343\times5=0.0021715$
$\therefore\log_{10}1005=\text{y}+\triangle\text{y}=3.0021715$

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