Question

1) se log de 2=0,301 e log 7=0,845, calcule log V140
2 3 4 5 estão no anexo ​

Answer 1

$\boxed{\begin{array}{l}\sf 1)~Se\,\ell og2=0,301\,e\,\ell og7=0,845\\\sf calcule\,\ell og\sqrt{140}.\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm \sqrt{140}=\sqrt{2\cdot7\cdot10}\\\\\rm \ell og\sqrt{140}=\ell og\sqrt{2\cdot7\cdot10}=\ell og(2\cdot7\cdot10)^{\frac{1}{2}}\\\rm \ell og\sqrt{140}=\dfrac{1}{2}\ell og(2\cdot7\cdot10)\\\\\rm \ell og\sqrt{140}=\dfrac{1}{2}(\ell og2+\ell og7+\ell og10)\\\\\rm\ell og\sqrt{140}=\dfrac{0,301+0,845+1}{2}\\\rm \ell og\sqrt{140}=\dfrac{2,146}{2}=1,073\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Propriedades\,operat\acute orias\,dos\,logaritmos}\\\rm 1- \ell og_c(a\cdot b)=\ell og_ca+\ell og_cb\\\sf 2-\rm \ell og_c\bigg(\dfrac{a}{b}\bigg)=\ell og_ca-\ell og_cb\\\\\sf 3-\rm \ell og_ca^n=n\ell og_ca\end{array}}$

$\large\boxed{\begin{array}{l}\sf 2)~Assinale\,a\,propriedade\,v\acute alida:\\\sf a) \ell og(a\cdot b)=\ell og a\cdot \ell ogb\\\sf b )\ell og(a+b)=\ell oga+\ell og b\\\sf c)\ell og m\cdot a=m\cdot \ell og a\\\sf d)\ell oga^m=m\cdot \ell oga\checkmark\\\rm nota: veja\,as\,propriedades\,operat\acute orias\\\rm dos\,logaritmos\,e\,conclua\\\rm que\,somente\,d\,\acute e\,verdadeira.\end{array}}$

$\large\boxed{\begin{array}{l}\sf 3)~Sendo\,\ell og2=0,301\,e\,\ell og7=0,845\\\sf qual\,ser\acute a\,o\,valor\,de\,\ell og28?\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm \ell og28=\ell og2^2\cdot7=\ell og2^2+\ell og7\\\rm \ell og28=2\ell og2+\ell og7\\\rm \ell og28=2\cdot0,301+0,845\\\rm \ell og28=0,602+0,845\\\rm \ell og28=1,447\end{array}}$

$\large\boxed{\begin{array}{l}\sf 4)\,Aplicando\,a\,propriedade\,dos\,log,desenvolva\\\sf \ell og\dfrac{ab}{c}\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm \ell og\bigg(\dfrac{ab}{c}\bigg)=\ell og(ab)-\ell ogc\\\\\rm \ell og\bigg(\dfrac{ab}{c}\bigg)=\ell oga+\ell ogb-\ell ogc\end{array}}$

$\large\boxed{\begin{array}{l}\sf5)~ Dados\,\ell og_ba=m\,e\,\ell og_bc=n, calcule~\ell og_ca.\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm \ell og_ca=\dfrac{\ell og_ba}{\ell og_bc}=\dfrac{m}{n}\end{array}}$