calculer la differentielle de y=log(x) teste: sachant que log(35)=1,54407, calculer log(3501) NB: on rappelle que (1/(log(10)))=log(e)=0,43429..


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Question Number 148312 by puissant last updated on 27/Jul/21

$c a l c u l e r l a d i f f e r e n t i e l l e d e$ $y = l o g (x)$ $t e s t e : s a c h a n t q u e l o g (35) = 1, 54407,$ $c a l c u l e r l o g (3501)$ $N B : o n r a p p e l l e q u e \frac{1}{l o g (10)} = l o g (e) = 0, 43429 . .$
	Answered by Olaf_Thorendsen last updated on 27/Jul/21

	$On a la relation : \log_{a} x = \frac{\ln x}{\ln_{a}}$ $En particulier \log x = \frac{\ln x}{\ln 10} (1)$ $Attention aux notations . Dans les$ $pays anglo - saxons, le logarithme$ $neperien \ln (base e) est souvent$ $note \log et le logarithme decimal \log$ $(base 10) est souvent note Log (avec$ $une majuscule) .$ $Dans notre cas, la notation \log dans$ $(1) est bien un logaritme decimal$ $(notation francaise donc) .$ $Et d \log (x) = d \frac{\ln x}{\ln 10} = \frac{1}{\ln 10} . \frac{d x}{x}$ $On en deduit une estimation de$ $l^{'} erreur (tres utile en physique) :$ $Δ y = \frac{1}{\ln 10} . \frac{Δ x}{x}$ $\log (3501) = \log (35, 01) + 2$ $Δ \log (35, 01) = 0, 43429 . \frac{0, 01}{35} \approx 1, 24 . 10^{- 4}$ $\log (3501) \approx \log (35) + Δ \log (35) + 2$ $\log (3501) \approx 1, 54407 + 1, 24 . 10^{- 4} + 2$ $\log (3501) \approx 3, 54419$
	Commented by puissant last updated on 27/Jul/21

	$m e r c i . .$
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