Question

log₁₀(10.01),Find approximate value.

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

$\sf{ The \: approximate \: value \: of \: \: \: log_{10}(10.01) }$

CALCULATION

$\sf{Let \: \: y = f(x) = log_{10}( x)}$

$\sf{Let \: us \: \: take \: \: \: x = 10 \: \: , \: \: \Delta x = 0.01}$

Now

$\sf{ f(x) }$

$\sf{ = log_{10}( x) }$

$\sf{ = log_{e}(x) \times log_{10}(e) }$

$\sf{ = log_{e}(x) \times 0.4343 \: \: ( \because \: log_{10}(e) = 0.4343 \: ) }$

$\sf{ \therefore \: \: f(x)= log_{e}(x) \times 0.4343 \: }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ f \: '(x)\: = \frac{0.4343}{x} }$

We know from application of derivatives that

$\sf{ f( x+ \Delta x) = f(x) + f \: ' (x)\:\Delta x }$

Putting the values we get

$\sf{ f(10 + 0.01) = f(10) + \: f \: ' (10) \times 0.01\: }$

$\displaystyle \: \implies \: \sf{ f(10.01) = log_{10}(10) + \: \frac{0.4343}{10} \times 0.01\: }$

$\displaystyle \: \implies \: \sf{ log_{10}(10.01) = 1 + 0.0004343\: }$

$\displaystyle \: \implies \: \sf{ log_{10}(10.01) = 1.0004343\: }$

RESULT

$\boxed{ \displaystyle \: \: \sf{ log_{10}(10.01) = 1.0004343\: \: \: }}$

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