Question

1) sendo log2=0,3 ;log3 =0,4 e log 5=0,7, calcule
2,3,4estao no anexo ​

Answer 1

$\large\boxed{\begin{array}{l}\rm 1)\,Sendo\,\ell og2=0,3;\ell og3=0,4\,e\ell og5=0,7\,,\,calcule\!:\\\rm a)\,\ell og50~~b)\ell og_345~~c)\ell og_92\\\rm d)\ell og_8600~~e)\ell og_53~~f)\ell og_615\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\boldsymbol{soluc_{\!\!,}\tilde ao\!:}}\\\rm a)~\sf \ell og_250=\dfrac{\ell og50}{\ell og2}=\dfrac{ \ell og(5\cdot10)}{\ell og2}\\\\\sf \ell og_250=\dfrac{\ell og5+\ell og10}{\ell og2}=\dfrac{0,7+1}{0,3}=\dfrac{1,7}{0,3}=\dfrac{17}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm b )~\sf \ell og_345=\dfrac{\ell og45}{\ell og3}=\dfrac{\ell og (5\cdot 3^2)}{\ell og3}=\dfrac{\ell og5+\ell og3^2}{\ell og3}\\\\\sf \ell og_345=\dfrac{ \ell og5+2\ell og3}{\ell og3}=\dfrac{0,7+2\cdot0,4}{0,4}=\dfrac{0,7+0,8}{0,4}\\\\\sf\ell og_345=\dfrac{1,5}{0,4}=\dfrac{15}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\rm c )~\sf\ell og_92=\dfrac{\ell og2}{\ell og9}=\dfrac{\ell og2}{\ell og3^2}=\dfrac{\ell og2}{2\ell og3}\\\\\sf \ell og_92=\dfrac{0,3}{2\cdot0,4}=\dfrac{0,3}{0,8}=\dfrac{3}{8}\end{array}}$

$\large\boxed{\begin{array}{l}\rm d)~\sf \ell og_8600=\dfrac{\ell og600}{\ell og8}=\dfrac{\ell og(2\cdot3\cdot10^2)}{\ell og2^3}\\\\\sf \ell og_8600=\dfrac{\ell og2+\ell og3+2\ell og10}{3\ell og2}\\\\\sf \ell og_8600=\dfrac{0,3+0,4+2\cdot1}{3\cdot0,3}=\dfrac{0,7+0,4+2}{0,9}\\\\\sf \ell og_8600=\dfrac{3,1}{0,9}=\dfrac{31}{9}\end{array}}$

$\large\boxed{\begin{array}{l}\rm e)~\sf \ell og_53=\dfrac{\ell og3}{\ell og5}=\dfrac{0,4}{0,7}=\dfrac{4}{7}\end{array}}$

$\large\boxed{\begin{array}{l}\rm f)~\sf \ell og_615=\dfrac{\ell og15}{\ell og6}=\dfrac{\ell og(3\cdot5)}{\ell og(2\cdot3)}\\\\\sf \ell og_615=\dfrac{\ell og3+\ell og5}{\ell og2+\ell og3}=\dfrac{0,4+0,7}{0,3+0,4}\\\\\sf \ell og_615=\dfrac{1,1}{0,7}=\dfrac{11}{7}\end{array}}$

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$\large\boxed{\begin{array}{l}\rm 2)\, Efetue\,o\,produto~~\ell og_32\cdot \ell og_25\cdot \ell og_53.\\\underline{\boldsymbol{soluc_{\!\!,}\tilde ao\!:}}\\\sf \ell og_32\cdot \ell og_25\cdot \ell og_53=\dfrac{\ell og2}{\ell og3}\cdot\dfrac{\ell og5}{\ell og2}\cdot\dfrac{\ell og3}{\ell og5}\\\rm organizando\,temos:\\\sf \ell og_32\cdot\ell og_25\cdot \ell og_53=\dfrac{\ell og2}{\ell og2}\cdot\dfrac{\ell og3}{\ell og3}\cdot\dfrac{\ell og5}{\ell go5}=1\end{array}}$

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$\large\boxed{\begin{array}{l}\rm 3)\,Dados\,\ell og_ba=m\,e\,\ell og_bc=n,calcule~\ell og_ca.\\\underline{\boldsymbol{soluc_{\!\!,}\tilde ao\!:}}\\\sf \ell og_ca=\dfrac{\ell og_ba}{\ell og_bc}\\\\\sf \ell og_ca=\dfrac{m}{n}\end{array}}$

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$\large\boxed{\begin{array}{l}\rm 4)\,Sendo\,\ell og_xa=6\,e\,\ell og_xb=4\,e\,\ell og_xc=2,\\\rm calcule:\\\rm a)~\ell og_c\sqrt{abc}~~~~~~~~~~~b)~\ell og_c(a^3\cdot b^2)\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\boldsymbol{soluc_{\!\!,}\tilde ao\!:}}\\\rm a)~\sf \ell og_c\sqrt{abc}=\dfrac{ \ell og_x\sqrt{abc}}{\ell og_xc}=\dfrac{ \ell og_x(abc)^{\frac{1}{2}}}{\ell og_xc}\\\\\sf \ell og_c\sqrt{abc}=\dfrac{\frac{1}{2}\cdot\ell og_x(abc)}{\ell og_xc}=\dfrac{\frac{1}{2}[\ell og_xa+\ell og_xb+\ell og_xc]}{\ell og_xc}\\\\\sf \ell og_c\sqrt{abc}=\dfrac{\frac{1}{2}[6+4+2]}{2}=\dfrac{\frac{1}{2}\cdot12}{2}=\dfrac{6}{2}\\\\\sf \ell og_c\sqrt{abc}=3\end{array}}$

$\large\boxed{\begin{array}{l}\rm b)~\sf \ell og_c(a^3\cdot b^2)=\ell og_ca^3+\ell og_cb^2\\\sf \ell og_c(a^3\cdot b^2)=3\ell og_ca+2\ell og_cb\\\sf \ell og_c(a^3\cdot b^2)=3\cdot\dfrac{\ell og_xa_}{\ell og_xc}+2\cdot\dfrac{ \ell og_xb}{\ell og_xc}\\\\\sf \ell og_c(a^3\cdot b^2)=3\cdot\dfrac{6}{2}+2\cdot\dfrac{4}{2}\\\\\sf\ell og_c(a^3\cdot b^2)=3\cdot3+2\cdot2\\\sf \ell og_c(a^3\cdot b^2)=9+4=13\end{array}}$

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