• Matéria: Matemática
  • Autor: Anônimo
  • Perguntado 3 anos atrás

51 pela definição de cologaritimo, calcule :​

Anexos:

Respostas

respondido por: CyberKirito
2

\large\boxed{\begin{array}{l}\underline{\rm D~\!\!efinic_{\!\!,}\tilde ao\,de\,cologaritmo}\\\sf co\ell\! og_ba=-\ell og_ba\end{array}}

\large\boxed{\begin{array}{l}\rm a)~\sf co\ell og_28=-\ell og_28\\\sf co\ell og_28=-\ell og_{2^1}2^3=-\dfrac{3}{1}=-3\\\\\rm b)~\sf co\ell og_3\bigg(\dfrac{1}{81}\bigg)=-\ell og_3\bigg(\dfrac{1}{81}\bigg)\\\\\sf co\ell og_3\bigg(\dfrac{1}{81}\bigg)=-\bigg(\ell og_31-\ell og_381\bigg)\\\\\sf co\ell og_3\bigg(\dfrac{1}{81}\bigg)=\ell og_381-\ell og1=4\end{array}}

\large\boxed{\begin{array}{l}\rm c)~\sf co\ell og_{10}0,001=-\ell og_{10}10^{-3}=3\ell og_{10}10=3\end{array}}

\large\boxed{\begin{array}{l}\rm d)~\sf colog_22\sqrt{2}=co\ell og_22^{\frac{3}{2}}\\\\\sf co\ell og_22\sqrt{2}=-\ell og_22^{\frac{3}{2}}=-\dfrac{3}{2}\ell og_22=-\dfrac{3}{2}\end{array}}


Anônimo: brigadoo
Josiane718: Cyber
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