Question

Find the approximate value of \[{{\log }_{10}}\left( 1002 \right)\], given that \[{{\log }_{10}}e=0.4343\].

Answer 1

Hint: Take \[y=f\left( x \right)={{\log }_{10}}x\], where x = 1000. Then \[x+\Delta x=1002\]. Find \[\Delta x\] which is dx. Now find the value of \[{{\log }_{10}}1000\]. Find the value of \[\dfrac{dy}{dx}\] for x = 1000 and find \[\Delta y\] or \[dy\]. Add y and \[\Delta y\] to get the value of \[{{\log }_{10}}\left( 1002 \right)\].

Complete step-by-step answer:
We need to find the approximate value of \[{{\log }_{10}}\left( 1002 \right)\].
Let us consider, \[y=f\left( x \right)={{\log }_{10}}x\].
Here the value is log to the base 10.
Let us assume, x = 1000.
From \[{{\log }_{10}}\left( 1002 \right)\], we can say that,
\[\Rightarrow x+\Delta x=1002\]
\[\Delta x\] represents dx, put x = 1000 in the above,
\[\begin{align}
  & 1000+\Delta x=1002 \\
 & \Rightarrow \Delta x=1002-1000=2 \\
\end{align}\]
Thus we got the value of dx = \[\Delta x=2\].
Let us find the log value of, x = 1000.
We took, \[y=f\left( x \right)={{\log }_{10}}x\].
\[\Rightarrow y={{\log }_{10}}1000={{\log }_{10}}{{\left( 10 \right)}^{3}}=3\]
Thus we got y = 3.
Now, \[y={{\log }_{10}}x=\dfrac{{{\log }_{e}}x}{{{\log }_{e}}10}\]
We have been given that, \[{{\log }_{10}}e=0.4343\].
Thus from the above we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{0.4343}{x}\]
\[\dfrac{dy}{dx}\] for x = 1000 is given as,
\[\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{x=1000}}=\dfrac{0.4343}{1000}=0.0004343\]
Now, \[\Delta y=dy\].
\[\therefore \Delta y=dy=\dfrac{dy}{dx}\times dx=0.004343\times 2=0.0008686\]
\[\therefore {{\log }_{10}}1002=y+\Delta y=3+0.0008686=3.0008686\]
Thus we got the approximate value of \[{{\log }_{10}}\left( 1002 \right)\] as \[3.0008686\].

Note: You may write \[{{\log }_{10}}1002\] as \[{{\log }_{10}}\left( 1000+2 \right)\], which might yield the wrong answer. Thus you get \[{{\log }_{10}}1000=3\] and find \[{{\log }_{10}}2=0.301\]. On adding you get 3.031, which is slightly different from 3.0008686.